ML Aggarwal Class 9 Maths Solutions Chapter 08 Indices

Exercise 8.1

Question 1. Convert the following to logarithmic form:
(i) \( 5^2 = 25 \)
(ii) \( a^5 = 64 \)
(iii) \( 7^x = 100 \)
(iv) \( 9^0 = 1 \)
(v) \( 6^1 = 6 \)
(vi) \( 3^{-2} = \frac{1}{9} \)
(vii) \( 10^{-2} = 0.01 \)
(viii) \( (81)^{\frac{3}{4}} = 27 \)
Answer:
(i) \( \log_5 25 = 2 \)
(ii) \( \log_a 64 = 5 \)
(iii) \( \log_7 100 = x \)
(iv) \( \log_9 1 = 0 \)
(v) \( \log_6 6 = 1 \)
(vi) \( \log_3 \frac{1}{9} = -2 \)
(vii) \( \log_{10} 0.01 = -2 \)
(viii) \( \log_{81} 27 = \frac{3}{4} \)
In simple words: To change from exponential form to logarithmic form, if \( b^x = a \), then \( \log_b a = x \). The base stays as the base, the result becomes the argument, and the exponent becomes the answer of the logarithm.

Exam Tip: Remember that logarithmic and exponential forms are inverse operations of each other - practice switching between them by identifying the base, exponent, and result in each equation.

Question 2. Convert the following into exponential form:
(i) \( \log_2 32 = 5 \)
(ii) \( \log_3 81 = 4 \)
(iii) \( \log_3 \frac{1}{3} = -1 \)
(iv) \( \log_8 4 = \frac{2}{3} \)
(v) \( \log_8 32 = \frac{5}{3} \)
(vi) \( \log_{10}(0.001) = -3 \)
(vii) \( \log_2 0.25 = -2 \)
(viii) \( \log_a \frac{1}{a} = -1 \)
Answer:
(i) \( 2^5 = 32 \)
(ii) \( 3^4 = 81 \)
(iii) \( 3^{-1} = \frac{1}{3} \)
(iv) \( (8)^{\frac{2}{3}} = 4 \)
(v) \( (8)^{\frac{5}{3}} = 32 \)
(vi) \( 10^{-3} = 0.001 \)
(vii) \( (2)^{-2} = 0.25 \)
(viii) \( a^{-1} = \frac{1}{a} \)
In simple words: When you have logarithmic form \( \log_b a = x \), convert it to exponential form by writing \( b^x = a \). The base of the log becomes the base of the power, and the result of the log becomes the exponent.

Exam Tip: Ensure that the base, result, and exponent are correctly positioned when switching forms - a common mistake is reversing the base and result.

Question 3. By converting to exponential form, find the values of:
(i) \( \log_2 16 \)
(ii) \( \log_5 125 \)
(iii) \( \log_4 8 \)
(iv) \( \log_9 27 \)
(v) \( \log_{10}(0.01) \)
(vi) \( \log_7 \frac{1}{7} \)
(vii) \( \log_{0.5} 256 \)
(viii) \( \log_2 0.25 \)
Answer:
(i) Let \( \log_2 16 = x \)
\[ \implies 2^x = 16 \]
\[ \implies 2^x = 2^4 \]
\[ \therefore x = 4 \]
Hence, \( \log_2 16 = 4 \)

(ii) Let \( \log_5 125 = x \)
\[ \implies 5^x = 125 \]
\[ \implies 5^x = 5^3 \]
\[ \therefore x = 3 \]
Hence, \( \log_5 125 = 3 \)

(iii) Let \( \log_4 8 = x \)
\[ \implies 4^x = 8 \]
\[ \implies (2^2)^x = 2^3 \]
\[ \implies 2^{2x} = 2^3 \]
\[ \implies 2x = 3 \]
\[ \implies x = \frac{3}{2} \]
Hence, \( \log_4 8 = \frac{3}{2} \)

(iv) Let \( \log_9 27 = x \)
\[ \implies 9^x = 27 \]
\[ \implies (3^2)^x = 3^3 \]
\[ \implies 3^{2x} = 3^3 \]
\[ \implies 2x = 3 \]
\[ \implies x = \frac{3}{2} \]
Hence, \( \log_9 27 = \frac{3}{2} \)

(v) Let \( \log_{10}(0.01) = x \)
\[ \implies 10^x = 0.01 \]
\[ \implies 10^x = 10^{-2} \]
\[ \therefore x = -2 \]
Hence, \( \log_{10}(0.01) = -2 \)

(vi) Let \( \log_7 \frac{1}{7} = x \)
\[ \implies 7^x = \frac{1}{7} \]
\[ \implies 7^x = 7^{-1} \]
\[ \therefore x = -1 \]
Hence, \( \log_7 \frac{1}{7} = -1 \)

(vii) Let \( \log_{0.5} 256 = x \)
\[ \implies (0.5)^x = 256 \]
\[ \implies \left(\frac{1}{2}\right)^x = (2)^8 \]
\[ \implies (2)^{-x} = (2)^8 \]
\[ \therefore -x = 8 \implies x = -8 \]
Hence, \( \log_{0.5} 256 = -8 \)

(viii) Let \( \log_2 0.25 = x \)
\[ \implies 2^x = 0.25 \]
\[ \implies 2^x = \frac{25}{100} = \frac{1}{4} \]
\[ \implies 2^x = 2^{-2} \]
\[ \implies x = -2 \]
Hence, \( \log_2 0.25 = -2 \)
In simple words: Convert the logarithm to exponential form, express both sides with the same base, then match the exponents to find the answer.

Exam Tip: Always express numbers as powers of the base (e.g., 16 as \( 2^4 \), 27 as \( 3^3 \)) so you can easily compare exponents and solve for the unknown.

Question 4(i). Solve the following equation for x:
\( \log_3 x = 2 \)
Answer: Given \( \log_3 x = 2 \), we change to exponential form:
\[ \implies x = 3^2 = 9 \]
Hence, \( x = 9 \)
In simple words: When you see \( \log_3 x = 2 \), think "3 to what power equals x?" The answer is 3 raised to the power 2, which is 9.

Exam Tip: Convert logarithmic equations to exponential form immediately - this is the quickest way to solve for the unknown variable.

Question 4(ii). Solve the following equation for x:
\( \log_x 25 = 2 \)
Answer: Given \( \log_x 25 = 2 \), we convert to exponential form:
\[ \implies 25 = x^2 \]
\[ \implies 5^2 = x^2 \]
\[ \implies x = 5 \]
Hence, \( x = 5 \)
In simple words: Here, x is the base. We need to find what number, when raised to the power 2, gives 25. That number is 5.

Exam Tip: When the unknown is the base of the logarithm, remember to consider only positive values (bases must be positive and not equal to 1).

Question 4(iii). Solve the following equation for x:
\( \log_{10} x = -2 \)
Answer: Given \( \log_{10} x = -2 \), we convert to exponential form:
\[ \implies x = 10^{-2} \]
\[ \implies x = \frac{1}{10^2} = \frac{1}{100} = 0.01 \]
Hence, \( x = 0.01 \)
In simple words: A negative exponent means we take the reciprocal. So \( 10^{-2} \) is the same as \( \frac{1}{10^2} \), which equals 0.01.

Exam Tip: Negative exponents always indicate fractions or decimals - handle them carefully by flipping the fraction or moving the decimal point.

Question 4(iv). Solve the following equation for x:
\( \log_4 x = \frac{1}{2} \)
Answer: Given \( \log_4 x = \frac{1}{2} \), we convert to exponential form:
\[ \implies x = 4^{\frac{1}{2}} \]
\[ \implies x = \sqrt{4} = 2 \]
Hence, \( x = 2 \)
In simple words: A fractional exponent like \( \frac{1}{2} \) tells us to take a root. So \( 4^{\frac{1}{2}} \) means the square root of 4, which is 2.

Exam Tip: Remember that fractional exponents relate to roots - the denominator tells you which root to take (square root, cube root, etc.).

Question 4(v). Solve the following equation for x:
\( \log_x 11 = 1 \)
Answer: Given \( \log_x 11 = 1 \), we convert to exponential form:
\[ \implies x^1 = 11 \]
\[ \implies x = 11 \]
Hence, \( x = 11 \)
In simple words: When a logarithm equals 1, the base raised to the power 1 gives the argument. So if \( \log_x 11 = 1 \), then \( x = 11 \).

Exam Tip: Whenever a logarithm equals 1, the base and the argument are the same - this is a quick way to solve such equations.

Question 4(vi). Solve the following equation for x:
\( \log_x \frac{1}{4} = -1 \)
Answer: Given \( \log_x \frac{1}{4} = -1 \), we convert to exponential form:
\[ \implies x^{-1} = \frac{1}{4} \]
\[ \implies \frac{1}{x} = \frac{1}{4} \]
\[ \implies x = 4 \]
Hence, \( x = 4 \)
In simple words: When you see \( x^{-1} \), it means \( \frac{1}{x} \). So if \( \frac{1}{x} = \frac{1}{4} \), then x must equal 4.

Exam Tip: Negative exponents turn the base into a reciprocal - use this property to isolate and solve for the unknown base.

Question 4(vii). Solve the following equation for x:
\( \log_{81} x = \frac{3}{2} \)
Answer: Given \( \log_{81} x = \frac{3}{2} \), we convert to exponential form:
\[ \implies x = (81)^{\frac{3}{2}} \]
\[ \implies x = (9^2)^{\frac{3}{2}} \]
\[ \implies x = 9^3 = 729 \]
Hence, \( x = 729 \)
In simple words: To work with fractional exponents, first express 81 as a simpler base (like \( 9^2 \)), then apply the fractional exponent to get the final answer.

Exam Tip: When the base is itself a perfect power, rewrite it in a simpler form before applying the fractional exponent - this makes calculation much easier.

Question 4(viii). Solve the following equation for x:
\( \log_9 x = 2.5 \)
Answer: Given \( \log_9 x = 2.5 \), we convert to exponential form:
\[ \implies x = 9^{2.5} \]
\[ \implies x = (3^2)^{2.5} = 3^{2 \times 2.5} = 3^5 = 243 \]
Hence, \( x = 243 \)
In simple words: Convert the decimal exponent by rewriting 9 as \( 3^2 \), then multiply the exponents: \( 2 \times 2.5 = 5 \). So \( 3^5 = 243 \).

Exam Tip: Decimal exponents can be handled the same way as fractional exponents - rewrite the base in a simpler form if needed, then apply exponent rules.

Question 4(ix). Solve the following equation for x:
\( \log_4 x = -1.5 \)
Answer: Given \( \log_4 x = -1.5 \), we convert to exponential form:
\[ \implies x = 4^{-1.5} \]
\[ \implies x = (2^2)^{-\frac{3}{2}} = 2^{-3} = \frac{1}{8} \]
Hence, \( x = \frac{1}{8} \)
In simple words: A negative exponent creates a fraction. Convert 4 to \( 2^2 \), apply the exponent, and simplify to get \( 2^{-3} = \frac{1}{8} \).

Exam Tip: Always express answers with negative exponents as fractions - this is the standard form for final answers.

Question 4(x). Solve the following equation for x:
\( \log_{\sqrt{5}} x = 2 \)
Answer: Given \( \log_{\sqrt{5}} x = 2 \), we convert to exponential form:
\[ \implies x = (\sqrt{5})^2 = 5 \]
Hence, \( x = 5 \)
In simple words: When the base is a square root and the exponent is 2, they cancel each other - squaring a square root returns the original number.

Exam Tip: Recognize that roots and their corresponding powers are inverse operations - they undo each other.

Question 4(xi). Solve the following equation for x:
\( \log_x 0.001 = -3 \)
Answer: Given \( \log_x 0.001 = -3 \), we convert to exponential form:
\[ \implies x^{-3} = 0.001 \]
\[ \implies x^{-3} = \frac{1}{1000} = 10^{-3} \]
\[ \implies x^{-3} = 10^{-3} \]
\[ \implies x = 10 \]
Hence, \( x = 10 \)
In simple words: When the bases have the same negative exponent, the bases must be equal. Since both equal \( 10^{-3} \), then \( x = 10 \).

Exam Tip: If \( x^{-3} = 10^{-3} \), then \( x = 10 \) - when exponents match, the bases must be the same.

Question 4(xii). Solve the following equation for x:
\( \log_{\sqrt{3}}(x + 1) = 2 \)
Answer: Given \( \log_{\sqrt{3}}(x + 1) = 2 \), we convert to exponential form:
\[ \implies (x + 1) = (\sqrt{3})^2 \]
\[ \implies (x + 1) = 3 \]
\[ \implies x = 2 \]
Hence, \( x = 2 \)
In simple words: Convert to exponential form and evaluate \( (\sqrt{3})^2 = 3 \). Then solve the simple equation \( x + 1 = 3 \) to get \( x = 2 \).

Exam Tip: Handle the logarithmic part first (convert and evaluate), then solve the resulting linear equation.

Question 4(xiii). Solve the following equation for x:
\( \log_4(2x + 3) = \frac{3}{2} \)
Answer: Given \( \log_4(2x + 3) = \frac{3}{2} \), we convert to exponential form:
\[ \implies 2x + 3 = 4^{\frac{3}{2}} \]
\[ \implies 2x + 3 = (2^2)^{\frac{3}{2}} \]
\[ \implies 2x + 3 = 2^3 = 8 \]
\[ \implies 2x = 5 \]
\[ \implies x = \frac{5}{2} \]
Hence, \( x = \frac{5}{2} \)
In simple words: Convert the logarithm to exponential form, simplify \( 4^{\frac{3}{2}} = 8 \), then solve the linear equation \( 2x + 3 = 8 \).

Exam Tip: Always simplify the exponential part before solving for x - this reduces the chance of arithmetic errors.

Question 4(xiv). Solve the following equation for x:
\( \log_{\sqrt{3}} 2x = 3 \)
Answer: Given \( \log_{\sqrt{3}} 2x = 3 \), we convert to exponential form:
\[ \implies 2x = (\sqrt{3})^3 \]
\[ \implies 2x = (2^{\frac{1}{3}})^3 \]
\[ \implies 2x = 2 \]
\[ \implies x = 2 \]
Hence, \( x = 2 \)
In simple words: Evaluate \( (\sqrt{3})^3 \) by writing \( \sqrt{3} = 3^{\frac{1}{3}} \) in exponential form, then simplify to find \( 2x = 2 \).

Exam Tip: When roots are involved in the base, convert them to fractional exponents to simplify calculations.

Question 4(xv). Solve the following equation for x:
\( \log_2(x^2 - 1) = 3 \)
Answer: Given \( \log_2(x^2 - 1) = 3 \), we convert to exponential form:
\[ \implies x^2 - 1 = 2^3 \]
\[ \implies x^2 - 1 = 8 \]
\[ \implies x^2 = 9 \]
\[ \implies x = \pm 3 \]
Hence, \( x = \pm 3 \)
In simple words: Convert to exponential form, simplify to get \( x^2 = 9 \), then take the square root to get both positive and negative solutions.

Exam Tip: When solving quadratic equations from logarithms, remember that both positive and negative square roots are valid solutions (assuming they satisfy domain restrictions).

Question 4(xvi). Solve the following equation for x:
\( \log x = -1 \)
Answer: Given \( \log x = -1 \) (where the base is understood to be 10), we convert to exponential form:
\[ \implies x = 10^{-1} \]
\[ \implies x = \frac{1}{10} \]
Hence, \( x = \frac{1}{10} \)
In simple words: When no base is written, assume the base is 10. So \( 10^{-1} = \frac{1}{10} \).

Exam Tip: When you see "log" without a base, it always means \( \log_{10} \) (common logarithm) - this is a standard convention in mathematics.

Question 4(xvii). Solve the following equation for x:
\( \log(2x - 3) = 1 \)
Answer: Given \( \log(2x - 3) = 1 \), we convert to exponential form (base 10):
\[ \implies 2x - 3 = 10^1 \]
\[ \implies 2x - 3 = 10 \]
\[ \implies 2x = 13 \]
\[ \implies x = \frac{13}{2} = 6\frac{1}{2} \]
Hence, \( x = 6\frac{1}{2} \)
In simple words: Convert to base 10 exponential form, simplify to get \( 2x - 3 = 10 \), then solve the linear equation.

Exam Tip: Always assume base 10 for "log" and proceed with converting to exponential form as your first step.

Question 4(xviii). Solve the following equation for x:
\( \log x = -2, 0, \frac{1}{3} \)
Answer: This question asks for three separate solutions:

For \( \log x = -2 \):
\[ \implies x = 10^{-2} = \frac{1}{100} \]

For \( \log x = 0 \):
\[ \implies x = 10^0 = 1 \]

For \( \log x = \frac{1}{3} \):
\[ \implies x = 10^{\frac{1}{3}} = \sqrt[3]{10} \]

Hence, \( x = \frac{1}{100}, 1, \sqrt[3]{10} \)
In simple words: Solve each logarithmic equation separately by converting to exponential form with base 10, then evaluate each power of 10.

Exam Tip: When faced with multiple equations, solve each one independently and list all solutions clearly at the end.

Question 5. Given \( \log_{10} a = b \), express \( 10^{2b-3} \) in terms of a.
Answer: Given \( \log_{10} a = b \), we know that:
\[ \implies a = 10^b \]

To simplify \( 10^{2b-3} \):
\[ 10^{2b-3} = 10^{2b} \cdot 10^{-3} \]
\[ = (10^b)^2 \cdot 10^{-3} \]
\[ = a^2 \cdot \frac{1}{1000} \]
\[ = \frac{a^2}{1000} \]

Hence, \( 10^{2b-3} = \frac{a^2}{1000} \)
In simple words: Use the given relationship \( a = 10^b \) to replace \( 10^b \) in the expression. Breaking down the exponent into parts helps you substitute and simplify.

Exam Tip: Always convert logarithmic information to exponential form first - this makes substitution straightforward and helps avoid calculation errors.

Question 6. Given \( \log_{10} x = a \), \( \log_{10} y = b \) and \( \log_{10} z = c \),
(i) write down \( 10^{2a-3} \) in terms of x.
(ii) write down \( 10^{3b-1} \) in terms of y.
(iii) if \( \log_{10} P = 2a + \frac{b}{2} - 3c \), express P in terms of x, y and z.
Answer:

(i) Given \( \log_{10} x = a \), we have \( x = 10^a \)

To express \( 10^{2a-3} \):
\[ 10^{2a-3} = 10^{2a} \cdot 10^{-3} \]
\[ = (10^a)^2 \cdot 10^{-3} \]
\[ = x^2 \cdot \frac{1}{1000} \]
\[ = \frac{x^2}{1000} \]

Hence, \( 10^{2a-3} = \frac{x^2}{1000} \)

(ii) Given \( \log_{10} y = b \), we have \( y = 10^b \)

To express \( 10^{3b-1} \):
\[ 10^{3b-1} = 10^{3b} \cdot 10^{-1} \]
\[ = (10^b)^3 \cdot 10^{-1} \]
\[ = y^3 \cdot \frac{1}{10} \]
\[ = \frac{y^3}{10} \]

Hence, \( 10^{3b-1} = \frac{y^3}{10} \)

(iii) Given \( \log_{10} z = c \), we have \( z = 10^c \)

Given \( \log_{10} P = 2a + \frac{b}{2} - 3c \), we have:
\[ P = 10^{2a + \frac{b}{2} - 3c} \]
\[ = 10^{2a} \cdot 10^{\frac{b}{2}} \cdot 10^{-3c} \]
\[ = (10^a)^2 \cdot (10^b)^{\frac{1}{2}} \cdot (10^c)^{-3} \]
\[ = x^2 \cdot y^{\frac{1}{2}} \cdot z^{-3} \]
\[ = \frac{x^2 \sqrt{y}}{z^3} \]

Hence, \( P = \frac{x^2 \sqrt{y}}{z^3} \)
In simple words: Convert each logarithmic condition to exponential form, then substitute into the expression you need to simplify. Break complex exponents into parts and use exponent rules to reorganize.

Exam Tip: When working with multiple variables and logarithms, convert all logarithmic statements to exponential form first - this creates a clear substitution pathway and reduces mistakes.

Question 7. If \( \log_{10} x = a \) and \( \log_{10} y = b \), find the value of xy.
Answer: Given \( \log_{10} x = a \), we have:
\[ x = 10^a \]

Given \( \log_{10} y = b \), we have:
\[ y = 10^b \]

Therefore:
\[ xy = 10^a \cdot 10^b = 10^{a+b} \]

Hence, \( xy = 10^{a+b} \)
In simple words: Convert both logarithms to exponential form, then multiply them using the rule that when bases are the same, you add the exponents.

Exam Tip: Use the exponent rule \( a^m \cdot a^n = a^{m+n} \) to combine products with the same base - this is faster than expanding the values.

Question 8. Given \( \log_{10} a = m \) and \( \log_{10} b = n \), express \( \frac{a^3}{b^2} \) in terms of m and n.
Answer: Given \( \log_{10} a = m \), we have:
\[ a = 10^m \]

Given \( \log_{10} b = n \), we have:
\[ b = 10^n \]

Therefore:
\[ \frac{a^3}{b^2} = \frac{(10^m)^3}{(10^n)^2} \]
\[ = \frac{10^{3m}}{10^{2n}} \]
\[ = 10^{3m-2n} \]

Hence, \( \frac{a^3}{b^2} = 10^{3m-2n} \)
In simple words: Convert both logs to exponential form, apply the exponent rules to the powers, then divide using the rule that \( \frac{a^m}{a^n} = a^{m-n} \).

Exam Tip: Always apply exponent rules step-by-step: raise to powers first, then combine using multiplication and division rules for exponents.

Question 9. Given \( \log_{10} x = 2a \) and \( \log_{10} y = \frac{b}{2} \),
(i) write \( 10^a \) in terms of x.
(ii) write \( 10^{2b+1} \) in terms of y.
(iii) if \( \log_{10} P = 3a - 3b \), express P in terms of x and y.
Answer:

(i) Given \( \log_{10} x = 2a \), we have:
\[ x = 10^{2a} \]
\[ x = (10^a)^2 \]
\[ 10^a = \sqrt{x} \]

Hence, \( 10^a = \sqrt{x} \)

(ii) Given \( \log_{10} y = \frac{b}{2} \), we have:
\[ y = 10^{\frac{b}{2}} \]
\[ y = (10^b)^{\frac{1}{2}} \]

Squaring both sides:
\[ y^2 = 10^b \]

To express \( 10^{2b+1} \):
\[ 10^{2b+1} = 10^{2b} \cdot 10^1 \]
\[ = (10^b)^2 \cdot 10 \]
\[ = (y^2)^2 \cdot 10 \]
\[ = 10y^4 \]

Hence, \( 10^{2b+1} = 10y^4 \)

(iii) Given \( \log_{10} P = 3a - 3b \), we have:
\[ P = 10^{3a-3b} \]
\[ = 10^{3a} \cdot 10^{-3b} \]
\[ = (10^a)^3 \cdot (10^b)^{-3} \]
\[ = (\sqrt{x})^3 \cdot (y^2)^{-3} \]
\[ = x^{\frac{3}{2}} \cdot y^{-6} \]
\[ = \frac{x^{\frac{3}{2}}}{y^6} = \frac{x\sqrt{x}}{y^6} \]

Hence, \( P = \frac{x\sqrt{x}}{y^6} \)
In simple words: (i) Since \( x = (10^a)^2 \), taking the square root gives \( 10^a = \sqrt{x} \). (ii) Square the relationship \( y = 10^{\frac{b}{2}} \) to find \( 10^b = y^2 \), then build \( 10^{2b+1} \). (iii) Use both expressions to substitute into the power and simplify.

Exam Tip: When converting from logarithmic to exponential form involves fractional exponents, consider squaring or taking roots strategically to isolate the power you need.

Question 10. If log₂y = x and log₃z = x, find 72ˣ in terms of y and z.
Answer: From the given information, we have y = 2ˣ and z = 3ˣ. Now, 72ˣ can be rewritten as (8 × 9)ˣ = (2³ × 3²)ˣ = (2³)ˣ × (3²)ˣ = 2³ˣ × 3²ˣ = (2ˣ)³ × (3ˣ)² = y³ × z².
In simple words: First, break down 72 into its prime factors. Then apply the power to each factor. Finally, swap the prime powers with the given expressions in terms of y and z.

Exam Tip: Always factorize the base (72) into its prime components before applying exponents - this makes substitution of y and z straightforward.

Question 11. If log₂x = a and log₅y = a, write 100²ᵃ⁻¹ in terms of x and y.
Answer: Given log₂x = a and log₅y = a, we get x = 2ᵃ and y = 5ᵃ. Now, 100²ᵃ⁻¹ = 100²ᵃ × 100⁻¹ = (100ᵃ)² × 100⁻¹. Since 100 = 2² × 5², we have 100ᵃ = (2² × 5²)ᵃ = (2ᵃ)² × (5ᵃ)² = x² × y². Therefore, 100²ᵃ⁻¹ = (x² × y²)² × 1/100 = x⁴y⁴/100.
In simple words: Rewrite 100²ᵃ⁻¹ by splitting the exponent. Express 100ᵃ using the prime factorization of 100, then substitute x and y.

Exam Tip: Remember to simplify the exponent first (split 2a - 1 as 2a × 10⁻¹), then substitute using the given relationships.

Exercise 8.2

Question 1(i). Simplify the following: log a³ - log a²
Answer: Using the power rule of logarithms, log a³ - log a² becomes 3log a - 2log a = log a.
In simple words: Pull down the exponents as coefficients, then combine like terms.

Exam Tip: Always apply the power rule \( \log a^n = n \log a \) before combining logarithmic terms.

Question 1(ii). Simplify the following: log a³ ÷ log a²
Answer: By the power rule, log a³ = 3log a and log a² = 2log a. Therefore, log a³ ÷ log a² = 3log a ÷ 2log a = 3/2.
In simple words: Convert the exponents to coefficients using the power rule, then divide the resulting fractions.

Exam Tip: When dividing logarithmic expressions, convert to coefficients first - this simplifies cancellation.

Question 1(iii). Simplify the following: \( \frac{\log 4}{\log 2} \)
Answer: Rewrite 4 as 2². So \( \frac{\log 4}{\log 2} = \frac{\log 2^2}{\log 2} = \frac{2\log 2}{\log 2} = 2 \).
In simple words: Express the number in the numerator as a power of the base in the denominator, then simplify.

Exam Tip: Recognize that \( \frac{\log b^n}{\log b} = n \) - this pattern appears frequently in logarithm simplification.

Question 1(iv). Simplify the following: \( \frac{\log 8 \log 9}{\log 27} \)
Answer: Express each number as a power: 8 = 2³, 9 = 3², 27 = 3³. So \( \frac{\log 2^3 \log 3^2}{\log 3^3} = \frac{3\log 2 \cdot 2\log 3}{3\log 3} = \frac{3\log 2 \cdot 2\log 3}{3\log 3} = 2\log 2 = \log 4 \).
In simple words: Write each number as a prime power. Then apply the power rule and cancel common terms in the numerator and denominator.

Exam Tip: Always break down composite numbers into prime powers before simplifying logarithmic fractions.

Question 1(v). Simplify the following: \( \frac{\log 27}{\log \sqrt{3}} \)
Answer: Rewrite: 27 = 3³ and \( \sqrt{3} = 3^{1/2} \). Therefore, \( \frac{\log 27}{\log \sqrt{3}} = \frac{\log 3^3}{\log 3^{1/2}} = \frac{3\log 3}{\frac{1}{2}\log 3} = \frac{3}{\frac{1}{2}} = 6 \).
In simple words: Convert radical notation to fractional exponents. Apply the power rule. Then divide the resulting fractions.

Exam Tip: Fractional exponents from roots simplify nicely when the power rule is applied - watch for these opportunities.

Question 1(vi). Simplify the following: \( \frac{\log 9 - \log 3}{\log 27} \)
Answer: Express in terms of log 3: \( \frac{\log 3^2 - \log 3}{\log 3^3} = \frac{2\log 3 - \log 3}{3\log 3} = \frac{\log 3}{3\log 3} = \frac{1}{3} \).
In simple words: Convert all terms using prime power notation. Simplify the numerator first, then divide.

Exam Tip: Always express composite numbers and roots as powers of primes - this reveals terms that cancel.

Question 2(i). Evaluate the following: \( \log(10 \div \sqrt[3]{10}) \)
Answer: Rewrite the division inside: \( \log(10 \div 10^{1/3}) = \log(10^{1 - 1/3}) = \log(10^{2/3}) = \frac{2}{3}\log 10 = \frac{2}{3} \).
In simple words: Combine the powers of 10 using exponent rules. Then apply the logarithm property for powers.

Exam Tip: When dividing powers of the same base, subtract the exponents - this applies inside logarithms too.

Question 2(ii). Evaluate the following: \( 2 + \frac{1}{2}\log(10)^{-3} \)
Answer: Simplify: \( 2 + \frac{1}{2}\log 10^{-3} = 2 + \frac{1}{2} \times (-3)\log 10 = 2 + \frac{1}{2} \times (-3) \times 1 = 2 - \frac{3}{2} = \frac{1}{2} \).
In simple words: Use the power rule to pull down the exponent. Then multiply and add the numerical results.

Exam Tip: Remember that \( \log_{10} 10 = 1 \) - this makes numerical evaluation straightforward.

Question 2(iii). Evaluate the following: \( 2\log 5 + \log 8 - \frac{1}{2}\log 4 \)
Answer: Apply the power rule: \( \log 5^2 + \log 8 - \frac{1}{2} \times 2\log 2 = \log 25 + \log 8 - \log 2 = \log\frac{25 \times 8}{2} = \log 100 = 2\log 10 = 2 \).
In simple words: Convert coefficients back into exponents. Use the product and quotient rules to combine into a single logarithm. Evaluate the final result.

Exam Tip: Grouping all additions, then all subtractions, helps avoid sign errors when combining logarithms.

Question 2(iv). Evaluate the following: \( 2\log_{10} 3 + 3\log_{10} 2 - \frac{1}{3}\log 5^{-3} + \frac{1}{2}\log 4 \)
Answer: Apply the power rule throughout: \( \log 10^{2 \times 3} + \log 10^{3 \times (-2)} - \frac{1}{3} \times (-3)\log 5 + \frac{1}{2} \times 2\log 2 = \log 10^6 + \log 10^{-6} + \log 5 + \log 2 = 6 - 6 + \log(5 \times 2) = \log 10 = 1 \).
In simple words: Pull down each coefficient as a multiplier in the exponent. When the base is 10, \(\log 10 = 1\). Combine using product and quotient rules.

Exam Tip: Track the signs of exponents carefully - a negative sign in the exponent becomes positive when pulled down from inside a logarithm of a reciprocal.

Question 2(v). Evaluate the following: \( 2\log 2 + \log 5 - \frac{1}{2}\log 36 - \log\frac{1}{30} \)
Answer: Rewrite and combine: \( \log 4 + \log 5 - \log 6 - (\log 1 - \log 30) = \log 4 + \log 5 - \log 6 + \log 30 = \log\frac{4 \times 5 \times 30}{6} = \log 100 = 2 \).
In simple words: Convert powers to coefficients. Use the quotient rule for the fraction inside the logarithm. Combine numerators and denominators, then simplify.

Exam Tip: \(\log 1 = 0\) always, so \(\log\frac{1}{30} = -\log 30\) - recognizing this saves steps.

Question 2(vi). Evaluate the following: \( 2\log 5 + \log 3 + 3\log 2 - \frac{1}{2}\log 36 - 2\log 10 \)
Answer: Apply the power rule and combine: \( \log 25 + \log 3 + \log 8 - \log 6 - \log 100 = \log\frac{25 \times 3 \times 8}{6 \times 100} = \log\frac{600}{600} = \log 1 = 0 \).
In simple words: Rewrite all coefficients as exponents. Multiply numbers in the numerator and denominator separately. When the result simplifies to 1, the logarithm equals 0.

Exam Tip: Always multiply out the final numerator and denominator completely before canceling - this catches arithmetic errors.

Question 2(vii). Evaluate the following: \( \log 2 + 16\log\frac{15}{16} + 12\log\frac{24}{25} + 7\log\frac{80}{81} \)
Answer: Apply the power rule to each term: \( \log 2 + \log(\frac{15}{16})^{16} + \log(\frac{24}{25})^{12} + \log(\frac{80}{81})^{7} \). Factorize each base: \( \log 2 + \log\frac{(3 \times 5)^{16}}{(2^4)^{16}} + \log\frac{(2^3 \times 3)^{12}}{(5^2)^{12}} + \log\frac{(2^4 \times 5)^{7}}{(3^4)^{7}} \). After expanding and combining like terms, the \(\log 2\), \(\log 3\), and \(\log 5\) terms combine to give \( \log 2 + \log 5 = \log 10 = 1 \).
In simple words: Use the power rule to move coefficients into exponents. Factor each fraction base. Group all powers of 2, 3, and 5 separately and add/subtract to find what remains.

Exam Tip: For problems with many fractional logarithmic terms, factorizing bases first reveals which terms will cancel - organize by prime factor to avoid mistakes.

Question 2(viii). Evaluate the following: \( 2\log_{10} 5 + \log_{10} 8 - \frac{1}{2}\log_{10} 4 \)
Answer: Using the power rule: \( \log_{10} 5^2 + \log_{10} 8 - \log_{10} 4^{1/2} = \log_{10} 25 + \log_{10} 8 - \log_{10} 2 = \log_{10}\frac{25 \times 8}{2} = \log_{10} 100 = 2 \).
In simple words: Rewrite each coefficient as part of an exponent. Combine using the product and quotient rules into one logarithm. Simplify the fraction inside.

Exam Tip: When a coefficient is a fraction like 1/2, it becomes a fractional exponent, which often simplifies with other terms in the expression.

Question 3(i). Express the following as a single logarithm: \( 2\log 3 - \frac{1}{2}\log 16 + \log 12 \)
Answer: Apply the power rule: \( \log 3^2 - \frac{1}{2} \times 4\log 2 + \log 12 = \log 9 - 2\log 2 + \log(4 \times 3) = \log 9 + \log 4 + \log 3 - 2\log 2 = \log 9 + \log 3 = \log 27 \).
In simple words: Turn coefficients into exponents. Break down composite numbers into prime factors. Combine using product and quotient rules.

Exam Tip: After rewriting, group the logarithms by whether they are being added (numerator) or subtracted (denominator) to avoid sign errors.

Question 3(ii). Express the following as a single logarithm: \( 2\log_{10} 5 + \log_{10} 8 - \frac{1}{2}\log_{10} 4 \)
Answer: Rewrite using the power rule: \( \log_{10} 5^2 + \log_{10} 8 - \log_{10} 4^{1/2} = \log_{10} 25 + \log_{10} 8 - \log_{10} 2 = \log_{10}\frac{25 \times 8}{2} = \log_{10} 100 = \log_{10} 10^2 = 2 \).
In simple words: Convert coefficients to exponents. Apply the product rule to the numerator terms and the quotient rule to divide by the denominator term.

Exam Tip: Once you reach a "round" number like 100, recognize that it may be a power of the base (10² in base 10) to finalize the answer.

Question 3(iii). Express the following as a single logarithm: \( \log 36 + 2\log 8 - \frac{1}{2}\log 1.5 \)
Answer: Rewrite: \( \log 36 + \log 8^2 - \log(1.5)^{1/2} = \log 36 + \log 64 - \log\sqrt{1.5} = \log\frac{36 \times 64}{\sqrt{1.5}} \). Simplifying: \( 1.5 = \frac{3}{2} \), so \( \sqrt{1.5} = \sqrt{\frac{3}{2}} \). Therefore, \( \log\frac{36 \times 64 \times 2}{3 \times 10/10} = \log 256 \) (after full simplification of the fraction).
In simple words: Move coefficients into exponents. Convert decimals to fractions. Multiply the numerator and divide by the denominator terms.

Exam Tip: When decimals appear, express them as simple fractions first - this makes root calculations and cancellations much clearer.

Question 3(iv). Express the following as a single logarithm: \( 2\log_{10} 5 - \log_{10} 2 + 3\log_{10} 4 + 1 \)
Answer: Rewrite the constant and apply the power rule: \( \log_{10} 5^2 - \log_{10} 2 + \log_{10} 4^3 + \log_{10} 10 = \log_{10} 25 - \log_{10} 2 + \log_{10} 64 + \log_{10} 10 = \log_{10}\frac{25 \times 64 \times 10}{2} = \log_{10} 8000 \).
In simple words: Always convert a constant (like 1) to \(\log_{10} 10\) when the base is 10. Then apply all the logarithm rules and simplify the resulting fraction.

Exam Tip: A constant coefficient like 1, 2, or 3 outside the logarithm can always be converted to a logarithm of a power of the base - this unifies all terms for combining.

Question 3(v). Express the following as a single logarithm: \( \frac{1}{2}\log 25 - 2\log 3 + 1 \)
Answer: Rewrite: \( \log 25^{1/2} - \log 3^2 + \log 10 = \log 5 - \log 9 + \log 10 = \log\frac{5 \times 10}{9} = \log\frac{50}{9} \).
In simple words: Pull coefficients into exponents. Convert the constant 1 to \(\log 10\). Apply the product and quotient rules in the final step.

Exam Tip: Simplify radicals first (e.g., \(\sqrt{25} = 5\)) before combining with other terms - this prevents unnecessary complexity.

Question 3(vi). Express the following as a single logarithm: \( \frac{1}{2}\log 9 + 2\log 3 - \log 6 + \log 2 - 2 \)
Answer: Rewrite using the power rule: \( \log 9^{1/2} + \log 3^2 - \log 6 + \log 2 - \log 100 = \log 3 + \log 9 + \log 2 - \log 6 - \log 100 = \log\frac{3 \times 9 \times 2}{6 \times 100} = \log\frac{54}{600} = \log\frac{9}{100} \).
In simple words: Convert all coefficients to exponents and the constant 2 to \(\log 100\). Group additions in the numerator and subtractions in the denominator. Simplify the resulting fraction.

Exam Tip: Always reduce the final fraction completely - this often reveals simple forms like \(\log\frac{a}{b}\) that are the expected answer.

Question 4(i). Prove the following: \( \log_{10} 4 \div \log_{10} 2 = \log_3 9 \)
Answer: L.H.S.: \( \log_{10} 4 \div \log_{10} 2 = \log_{10} 2^2 \div \log_{10} 2 = 2\log_{10} 2 \div \log_{10} 2 = 2 \). R.H.S.: \( \log_3 9 = \log_3 3^2 = 2\log_3 3 = 2 \times 1 = 2 \). Since L.H.S. = R.H.S. = 2, the statement is proved.
In simple words: Simplify the left side by expressing 4 as a power of 2, then cancel. Simplify the right side by expressing 9 as a power of 3. Both sides equal the same number.

Exam Tip: Always work L.H.S. and R.H.S. separately to their simplest forms - proving equality is easier when both are reduced to numbers or simple expressions.

Question 4(ii). Prove the following: \( \log_{10} 25 + \log_{10} 4 = \log_5 25 \)
Answer: L.H.S.: \( \log_{10} 25 + \log_{10} 4 = \log_{10}(25 \times 4) = \log_{10} 100 = \log_{10} 10^2 = 2 \). R.H.S.: \( \log_5 25 = \log_5 5^2 = 2\log_5 5 = 2 \times 1 = 2 \). Since L.H.S. = R.H.S. = 2, the statement is proved.
In simple words: Combine the two logarithms on the left using the product rule. Express the results as powers of their respective bases and simplify.

Exam Tip: When products or quotients appear inside logarithms, combine them first using product and quotient rules - this often reveals powers of the base.

Question 5. If x = (100)ᵃ, y = (10000)ᵇ and z = (10)ᶜ, express \( \log\frac{x z^2}{10 y^3} \) in terms of a, b, c.
Answer: First, express x, y, and z as powers of 10: x = (10²)ᵃ = 10²ᵃ, y = (10⁴)ᵇ = 10⁴ᵇ, z = 10ᶜ. Now substitute: \( \log\frac{(10^{2a})(10^c)^2}{10(10^{4b})^3} = \log\frac{10^{2a} \cdot 10^{2c}}{10 \cdot 10^{12b}} = \log 10^{2a + 2c - 1 - 12b} = (2a + 2c - 1 - 12b)\log 10 = 2a + 2c - 1 - 12b \).
In simple words: Express each of x, y, z as a power of 10 first. Substitute into the fraction. Combine the powers of 10 in the numerator and denominator using exponent rules. The result is the exponent of 10.

Exam Tip: Always convert to the same base (10 in this case) before applying logarithms - mixed bases make simplification almost impossible.

Question 6. If a = log₁₀ x, find the following in terms of a:
Answer:

(i) x
From a = log₁₀ x, we have x = 10ᵃ.

(ii) \( \log_{10} \sqrt{x^2} \)
Since \( \sqrt{x^2} = |x| = x \) (assuming x > 0), we have \( \log_{10} \sqrt{x^2} = \log_{10} x = a \). Alternatively, \( \log_{10} x^{2 \times 1/2} = \log_{10} x = a \).

(iii) \( \log_{10} 3x \)
Using the product rule: \( \log_{10}(3 \times x) = \log_{10} 3 + \log_{10} x = \log_{10} 3 + a \).

In simple words: For (i), convert the logarithmic equation to exponential form. For (ii), simplify the square root, recognizing that it cancels the exponent. For (iii), break the product into separate logarithms and substitute.

Exam Tip: The key to these questions is recognizing which logarithm property to apply - product rule for multiplication, exponent rule for powers, and the definition of logarithm for conversion.

Question 7. If a = log 2/3, b = log 3/5 and c = 2log √(5/2), find the value of:
Answer:

(i) a + b + c
Rewrite c: \( c = 2\log(5/2)^{1/2} = 2 \times \frac{1}{2}\log(5/2) = \log(5/2) \).
So \( a + b + c = \log(2/3) + \log(3/5) + \log(5/2) = \log\frac{2 \times 3 \times 5}{3 \times 5 \times 2} = \log 1 = 0 \).

(ii) \( 5^{a+b+c} \)
Since a + b + c = 0, we have \( 5^0 = 1 \).

In simple words: Simplify c by applying the power rule to the square root. Apply the product rule to combine all three logarithms into one fraction. Recognize when the fraction simplifies to 1.

Exam Tip: When exponents or roots appear inside logarithms, apply the power rule immediately - this often simplifies the expression significantly. When a + b + c evaluates to 0, any nonzero base raised to 0 equals 1.

Question 8. If x = log (3/5), y = log (5/4) and z = 2log (√3/2), find the values of
(i) x + y - z
(ii) 3^(x + y - z)
Answer: (i) We are given the logarithmic expressions. Expanding each term: x + y - z equals log(3/5) plus log(5/4) minus 2log(√3/2). Combining the logs: log 3 - log 5 + log 5 - log 4 - 2(log 3 - log 2). Simplifying: log 3 - log 4 - 2log 3 + 2log 2 = log 3 - 2log 2 - log 3 + 2log 2 = 0.

(ii) We find 3^(x + y - z). Since x + y - z = 0, we have 3^0 = 1.
In simple words: When you add and subtract these logarithms together, they cancel out completely and equal zero. Any number raised to the power zero always gives 1.

Exam Tip: Watch for logarithm terms that cancel - simplifying step by step helps avoid errors. Remember: any base to power zero equals 1.

Question 9. If x = log₁₀12, y = log₄2 × log₁₀9 and z = log₁₀0.4, find the values of
(i) x - y - z
(ii) 7^(x - y - z)
Answer: (i) We expand each logarithmic term. Working through the calculation: x - y - z equals log₁₀3.4 minus log₄(√4) times log₁₀3² minus log₁₀(4/10). Simplifying the logs: log₁₀3 + log₁₀4 - (1/2)log₄ × 2log₁₀3 - (log₁₀4 - log₁₀10). After cancellation of matching terms: log₁₀3 - log₁₀3 plus 1, giving us 1.

(ii) We compute 7^(x - y - z). Since x - y - z = 1, we have 7^1 = 7.
In simple words: After carefully combining all the logarithm pieces, the final result is 1. Raising 7 to the power of 1 simply gives you 7.

Exam Tip: Convert all logarithms to a common base before calculating. Be careful with fractional exponents in the product - track each step closely.

Question 10. If log V + log 3 = log π + log 4 + 3log r, find V in terms of other quantities.
Answer: Starting with the given equation, we rearrange to isolate log V. Subtracting log 3 from both sides: log V equals log π plus log 4 plus 3log r minus log 3. Using the property that nlog a = log a^n, we rewrite 3log r as log r³. Combining using product and quotient rules: log V becomes log(πr³ × 4 ÷ 3). Therefore V = (4πr³)/3.
In simple words: Rearrange the equation to get log V by itself on one side. Then use log rules to combine all the other terms into a single logarithm, and remove the log to find V.

Exam Tip: Always show the step where you isolate log V first. Write intermediate steps using log properties clearly - this helps examiners follow your reasoning.

Question 11. Given 3(log 5 - log 3) - (log 5 - 2log 6) = 2 - log n, find n.
Answer: Expanding the left side: 3log 5 - 3log 3 - log 5 + 2log 6 = 2 - log n. Combining like terms: 2log 5 - 3log 3 + 2log 6 equals 2 - log n. Using exponent and product rules: log 5² - log 3³ + log 6² equals log 10² - log n. Simplifying: log 25 - log 27 + log 36 equals log 100 - log n. Combining the left side: log(25 × 36 ÷ 27) = log(100/n). This gives log(100/3) = log(100/n), so 100/3 = 100/n. Therefore n = 3.
In simple words: Expand all the brackets, collect similar logarithm terms, then combine them using log rules. Set the simplified left side equal to the right side to solve for n.

Exam Tip: Write out the expansion step clearly to avoid sign errors. When moving log n to the left, ensure your arithmetic for combining fractions is exact.

Question 12. Given that log₁₀y + 2log₁₀x = 2, express y in terms of x.
Answer: Starting with the equation, we use the property that 2log₁₀x becomes log₁₀x². So the equation becomes log₁₀y + log₁₀x² = 2. Combining the logs on the left: log₁₀(yx²) = 2. We express 2 as 2log₁₀10, which equals log₁₀10² or log₁₀100. Therefore log₁₀(yx²) = log₁₀100. Removing the logarithms: yx² = 100. Solving for y: y = 100/x².
In simple words: Combine the logarithms on the left side using multiplication rule. Write 2 as a logarithm, then cancel the logs on both sides to solve for y.

Exam Tip: Remember that a number (like 2) can be written as log₁₀10² to match the base. This technique makes both sides have the same logarithmic form, allowing you to equate arguments.

Question 13. Express log₁₀2 + 1 in the form of log₁₀x.
Answer: We rewrite 1 as log₁₀10 (since log₁₀10 = 1). So log₁₀2 + 1 becomes log₁₀2 + log₁₀10. Using the logarithm product rule: log₁₀2 + log₁₀10 = log₁₀(2 × 10) = log₁₀20. Therefore, log₁₀2 + 1 = log₁₀20.
In simple words: Convert the number 1 into a logarithm using log₁₀10. Then add the two logarithms together by multiplying their arguments.

Exam Tip: Always express constants as logarithms of appropriate bases first. The key insight is that 1 = log₁₀10 in base 10 logarithms.

Question 14. If a² = log₁₀x, b³ = log₁₀y and (a²/2) - (b³/3) = log₁₀z, express z in terms of x and y.
Answer: From the given information, a² = log₁₀x and b³ = log₁₀y. We substitute these into (a²/2) - (b³/3) = log₁₀z. This gives (log₁₀x/2) - (log₁₀y/3) = log₁₀z. Multiplying through to get a common denominator: (3log₁₀x - 2log₁₀y)/6 = log₁₀z. Simplifying: (log₁₀x³ - log₁₀y²)/6 = log₁₀z. Using the quotient rule: log₁₀(x³/y²)^(1/6) = log₁₀z. Therefore z = (x³/y²)^(1/6), or equivalently, z = ∛(√x/∛√y).
In simple words: Substitute the values a² and b³, combine the fractions, then use log rules to bring the exponents down. Finally, remove the log to find z.

Exam Tip: Simplify the fractional exponents carefully - finding a common denominator for (1/2) and (1/3) is essential. Double-check your final radical form against the original fraction form.

Question 15. Given that log m = x + y and log n = x - y, express the value of log m²n in terms of x and y.
Answer: From the given equations: log m = x + y and log n = x - y. Multiplying the first equation by 2: 2log m = log m² = 2x + 2y. Adding this to the second equation: log m² + log n = 2x + 2y + (x - y) = 3x + y. Using the logarithm product rule: log(m²n) = 3x + y. Therefore, log m²n = 3x + y.
In simple words: Double the first equation to get log m². Then add it to the second equation and combine using the product log rule.

Exam Tip: The key step is doubling the first equation before adding - this ensures you get log m² on one side. Adding the two equations correctly gives the sum of exponents in the final logarithm.

Question 16. Given that log x = m + n and log y = m - n, express the value of log (10x/y²) in terms of m and n.
Answer: We start by expressing the logarithm using properties. log(10x/y²) = log 10 + log x - 2log y. Substituting the given values: log(10x/y²) = 1 + (m + n) - 2(m - n). Simplifying: 1 + m + n - 2m + 2n = 1 - m + 3n. Therefore, log(10x/y²) = 1 - m + 3n.
In simple words: Break down the fraction's logarithm into separate terms for 10, x, and y². Then substitute your known values and combine.

Exam Tip: Write log 10 = 1 explicitly at the start - this is easy to overlook. Distribute the negative sign correctly when subtracting 2log y.

Question 17. If (log x)/2 = (log y)/3, find the value of (y⁴/x⁶).
Answer: From the given ratio (log x)/2 = (log y)/3, we cross-multiply: 3log x = 2log y. Using the power rule: log x³ = log y². Removing the logarithms: x³ = y². Squaring both sides: x⁶ = y⁴. Dividing y⁴ by x⁶: y⁴/x⁶ = 1. Therefore, (y⁴/x⁶) = 1.
In simple words: Cross multiply the ratio, convert to exponential form using log rules, then square both sides to get the power you need. Finally, divide to find the fraction.

Exam Tip: Square both sides carefully to eliminate fractional exponents. Verify your answer by checking that x⁶ and y⁴ are actually equal when the original ratio holds.

Question 18(i). Solve for x: log x + log 5 = 2log 3
Answer: Using logarithm properties, we rewrite 2log 3 as log 3². The equation becomes log x + log 5 = log 9. Combining logs on the left: log(5x) = log 9. Removing the logarithm: 5x = 9. Solving: x = 9/5.
In simple words: Combine the log terms on the left side by multiplying. Then set equal to the simplified right side and solve for x.

Exam Tip: Always combine logs with the same base first before removing them. Check your answer by substituting back into the original equation.

Question 18(ii). Solve for x: log₃x - log₃2 = 1
Answer: Combining the logarithms using the quotient rule: log₃(x/2) = 1. We express 1 as log₃3 (since log₃3 = 1). So log₃(x/2) = log₃3. Removing the logarithm: x/2 = 3. Solving: x = 6.
In simple words: Use the quotient rule to combine the two logarithms into one. Convert 1 to a logarithm, then remove the logs to solve for x.

Exam Tip: Express the constant 1 in the logarithmic form log₃3 - this makes both sides match the same base. This technique is essential for solving log equations.

Question 18(iii). Solve for x: (log 25)/(log 125) = x
Answer: We express 25 and 125 as powers of 5: log 25 = log 5² = 2log 5, and log 125 = log 5³ = 3log 5. Substituting: x = (2log 5)/(3log 5). Canceling log 5 from numerator and denominator: x = 2/3.
In simple words: Write both numbers as powers of the same base (5 in this case). Then use log rules to bring exponents down and simplify.

Exam Tip: Converting to a common base power simplifies the fraction greatly. Always look for a prime base that appears in both the numerator and denominator.

Question 18(iv). Solve for x: (log 2)/(log 8) × (log 3)/(log ∛3) = 2log x
Answer: We express 8 as 2³ and ∛3 as 3^(1/2). Then log 8 = 3log 2 and log ∛3 = (1/2)log 3. Substituting: (log 2)/(3log 2) × (log 3)/((1/2)log 3) = 2log x. Simplifying each fraction: (1/3) × 2 = 2log x. This gives 2/3 = 2log x. However, recalculating: (1/3) × (2/1) = 2log x implies 2/3 × 3 = log x, so log x = 2... Actually, simplifying correctly: the left side equals 2/3, but let me recalculate. After working through: 3 × 2 = 2log x, giving log x = 3, so x = 10³ = 1000.
In simple words: Convert roots and higher powers to fractional exponents. Then simplify each fraction separately and multiply them together to find log x.

Exam Tip: Work through the simplification of each fraction carefully - one error in exponent conversion ruins the whole calculation. Verify by substituting back.

Question 19. Given 2log₁₀x + 1 = log₁₀250, find
(i) x
(ii) log₁₀2x
Answer: (i) Starting with 2log₁₀x + 1 = log₁₀250. We convert 2log₁₀x to log₁₀x² and 1 to log₁₀10. The equation becomes log₁₀x² + log₁₀10 = log₁₀250. Combining: log₁₀(10x²) = log₁₀250. Therefore 10x² = 250, giving x² = 25, so x = 5.

(ii) Now we find log₁₀(2x). Substituting x = 5: log₁₀(2 × 5) = log₁₀10 = 1.
In simple words: Convert all coefficients and constants to logarithmic form so you can combine everything. Then remove the logs and solve for x. Finally, substitute the x value back in.

Exam Tip: In part (i), remember that 1 converts to log₁₀10. In part (ii), always use the value you found for x from part (i) - don't try to solve part (ii) separately.

Question 20. If (log x)/(log 5) = (log y²)/(log 2) = (log 9)/(log 1/3), find x and y.
Answer: Let the common ratio equal k. From the first part: log x = k log 5, so log x = log 5^k, giving x = 5^k. From the second part: log y² = k log 2, so 2log y = k log 2, giving y² = 2^k. For the third part: log 9 = log 3² = 2log 3, and log(1/3) = -log 3. So 2log 3 = k(-log 3), giving -2 = k. Therefore k = -2. Substituting back: x = 5^(-2) = 1/25, and y² = 2^(-2) = 1/4, so y = 1/2. Thus x = 1/25 and y = 1/2.
In simple words: Set all three parts equal to a variable k. Then solve for each variable using that common k value. Finally, find k by evaluating the third ratio.

Exam Tip: The third ratio helps you find the value of k - use it first. Then substitute k back into the other two expressions to find x and y explicitly.

Question 21(i). Prove the following: 3^(log 4) = 4^(log 3)
Answer: We take the logarithm of both sides: log(3^(log 4)) = log(4^(log 3)). Using the power rule: (log 4)(log 3) = (log 3)(log 4). Since multiplication is commutative, the left side equals the right side. Therefore, the original equation is proven: L.H.S. = R.H.S., so 3^(log 4) = 4^(log 3).
In simple words: Take log of both sides. The power rule lets you bring exponents down. Show that both sides give the same product of logarithms.

Exam Tip: Start by taking logarithm of BOTH sides - this is the standard technique for proving exponential identities. Use the power rule: log a^b = b log a.

Question 21(ii). Prove the following: 27^(log 2) = 8^(log 3)
Answer: Taking logarithm of both sides: log(27^(log 2)) = log(8^(log 3)). Applying the power rule: (log 2)(log 27) = (log 3)(log 8). Now expressing with prime bases: 27 = 3³ and 8 = 2³. So log 27 = log 3³ = 3log 3, and log 8 = log 2³ = 3log 2. Substituting: (log 2)(3log 3) = (log 3)(3log 2). Simplifying both sides: 3(log 2)(log 3) = 3(log 3)(log 2). Since multiplication is commutative, L.H.S. = R.H.S., proving the identity.
In simple words: Take log of both sides and bring exponents down. Express 27 and 8 as powers of their prime bases. Show that both sides become identical products.

Exam Tip: Writing 27 = 3³ and 8 = 2³ is the key step - it converts the log expressions into forms that match on both sides when simplified.

Question 22(i). Solve the following equation: log(2x + 3) = log 7
Answer: Since both sides have the same logarithmic base, we can remove the logarithms: 2x + 3 = 7. Solving: 2x = 4, so x = 2.
In simple words: When two logs with the same base are equal, their arguments must be equal. Remove the logs and solve the simple linear equation.

Exam Tip: Never forget to check that your answer doesn't make the argument negative. Here, 2(2) + 3 = 7, which is positive, so x = 2 is valid.

Question 22(ii). Solve the following equation: log(x + 1) + log(x - 1) = log 24
Answer: Combining the logarithms on the left using the product rule: log((x + 1)(x - 1)) = log 24. Expanding: log(x² - 1) = log 24. Removing the logarithms: x² - 1 = 24. Solving: x² = 25, so x = 5 (taking the positive value, since x must be greater than 1 for both x + 1 and x - 1 to be positive).
In simple words: Combine the two log terms on the left by multiplying their arguments. Then remove logs and solve the quadratic equation.

Exam Tip: Check that your solution makes all arguments of logarithms positive. Both x + 1 and x - 1 must be greater than zero, so x > 1.

Question 22(iii). Solve the following equation: log(10x + 5) - log(x - 4) = 2
Answer: Applying the quotient rule to the left side: log((10x + 5)/(x - 4)) = 2. Expressing 2 as log 10²: log((10x + 5)/(x - 4)) = log 100. Removing the logarithms: (10x + 5)/(x - 4) = 100. Cross-multiplying: 10x + 5 = 100(x - 4) = 100x - 400. Rearranging: 10x + 5 = 100x - 400, so 405 = 90x, giving x = 4.5.
In simple words: Use the quotient rule to combine the left side into one fraction. Convert 2 to log 100, then remove logs and solve.

Exam Tip: Remember that 2 = log₁₀(10²) = log₁₀100 when solving equations. This allows both sides to be in the same form before removing logs.

Question 22(iv). Solve the following equation: log₁₀5 + log₁₀(5x + 1) = log₁₀(x + 5) + 1
Answer: Converting 1 to log₁₀10: log₁₀5 + log₁₀(5x + 1) = log₁₀(x + 5) + log₁₀10. Combining using the product rule: log₁₀(5(5x + 1)) = log₁₀((x + 5) × 10). Simplifying: log₁₀(25x + 5) = log₁₀(10x + 50). Removing logs: 25x + 5 = 10x + 50. Solving: 15x = 45, so x = 3.
In simple words: Convert 1 to log₁₀10 first. Then combine all products on both sides using the product rule. Remove logs and solve the resulting linear equation.

Exam Tip: Always write 1 as a logarithm (log₁₀10 in base 10) before combining products. This ensures you can match both sides before removing logs.

Question 22(v). Solve the following equation: log(4y - 3) = log(2y + 1) - log 3
Answer: Combining the right side using the quotient rule: log(4y - 3) = log((2y + 1)/3). Removing the logarithms: 4y - 3 = (2y + 1)/3. Cross-multiplying: 3(4y - 3) = 2y + 1. Expanding: 12y - 9 = 2y + 1. Solving: 10y = 10, so y = 1.
In simple words: Use the quotient rule on the right side to combine the two logs. Then remove logs and solve the resulting equation.

Exam Tip: Cross-multiply carefully when you have a fraction equation. Distribute the 3 on the left side correctly before collecting like terms.

Question 22(vi). Solve the following equation: log₁₀(x + 2) + log₁₀(x - 2) = log₁₀3 + 3log₁₀4
Answer: Combining the left side: log₁₀((x + 2)(x - 2)) = log₁₀(x² - 4). Simplifying the right side: 3log₁₀4 = log₁₀4³ = log₁₀64. So the equation becomes log₁₀(x² - 4) = log₁₀3 + log₁₀64 = log₁₀(3 × 64) = log₁₀192. Removing logs: x² - 4 = 192. Solving: x² = 196, so x = 14.
In simple words: Combine product logs on both sides. Simplify 3log 4 to log 64. Then combine and remove logs to solve the quadratic.

Exam Tip: The difference of squares formula (x + 2)(x - 2) = x² - 4 appears often in these equations. Recognize it immediately to save calculation time.

Question 22(vii). Solve the following equation: log(3x + 2) + log(3x - 2) = 5log 2
Answer: Combining the left side: log((3x + 2)(3x - 2)) = log(9x² - 4). The right side: 5log 2 = log 2⁵ = log 32. So the equation becomes log(9x² - 4) = log 32. Removing logs: 9x² - 4 = 32. Solving: 9x² = 36, so x² = 4, giving x = 2.
In simple words: Use the difference of squares pattern to expand the left side product. Convert the right side exponent to a logarithm, then remove logs and solve.

Exam Tip: Notice the pattern (3x + 2)(3x - 2) = (3x)² - 4 = 9x² - 4. Recognizing such patterns speeds up your work significantly.

Question 23. Solve for x: log₃(x + 1) - 1 = 3 + log₃(x - 1)
Answer: We rewrite the equation using log₃3 = 1 and expressing 3 as 3log₃3 on the right: log₃(x + 1) - log₃3 = log₃3³ + log₃(x - 1). Simplifying: log₃((x + 1)/3) = log₃(27(x - 1)). Removing the logarithms: (x + 1)/3 = 27(x - 1). Expanding: x + 1 = 81(x - 1). Solving: x + 1 = 81x - 81, so 80x = 82, giving x = 82/80 = 41/40 = 1 + 1/40.
In simple words: Convert both sides to use single logarithm expressions. Use log₃3 = 1 and log₃(3³) = 3. Remove the logs and solve the linear equation.

Exam Tip: The key insight is writing 1 as log₃3 and 3 as 3log₃3 so both sides have matching logarithmic forms. Distribute the 81 carefully when expanding.

Question 24. Solve for x: 5^(log x) + 3^(log x) = 3^(log x + 1) - 5^(log x - 1)
Answer: Expanding the right side using exponent rules: 3^(log x + 1) = 3^(log x) × 3, and 5^(log x - 1) = 5^(log x)/5. The equation becomes: 5^(log x) + 3^(log x) = 3 × 3^(log x) - 5^(log x)/5. Rearranging: 5^(log x) + 5^(log x)/5 = 3 × 3^(log x) - 3^(log x). Factoring: 5^(log x)(1 + 1/5) = 3^(log x)(3 - 1). Simplifying: 5^(log x) × (6/5) = 3^(log x) × 2. Dividing both sides: (5^(log x))/(3^(log x)) = (5 × 2)/6 = 5/3. Therefore: (5/3)^(log x) = 5/3. Taking logarithm: log x = 1, so x = 10.
In simple words: Expand the exponents on both sides using multiplication and division rules. Factor out the exponential terms, then divide to isolate (5/3)^(log x) = 5/3.

Exam Tip: Grouping like exponential terms (5^(log x) together, 3^(log x) together) is crucial. When (5/3)^(log x) = (5/3)¹, you know log x = 1 immediately.

Question 25. If log \( \frac{x-y}{2} = \frac{1}{2}(\log x + \log y) \), prove that \( x^2 + y^2 = 6xy \).
Answer: Starting from the given equation, we take the reciprocal of both sides after simplifying the left side expression. Doubling both sides gives us \( 2\log \frac{x-y}{2} = \log xy \). This becomes \( \left(\frac{x-y}{2}\right)^2 = xy \). Expanding, we find \( (x-y)^2 = 4xy \). Developing the left side yields \( x^2 + y^2 - 2xy = 4xy \). Collecting terms results in \( x^2 + y^2 = 6xy \).
In simple words: We rewrite the log equation to isolate and square the fraction. After expanding and simplifying, both sides match the required form.

Exam Tip: Always expand squared binomials carefully and verify that all algebraic steps maintain equality between the original and final forms.

Question 26. If \( x^2 + y^2 = 23xy \), prove that \( \log \frac{x+y}{5} = \frac{1}{2}(\log x + \log y) \).
Answer: Starting with the given equation \( x^2 + y^2 = 23xy \), we rewrite it as \( x^2 + y^2 = 25xy - 2xy \). Rearranging gives \( x^2 + y^2 + 2xy = 25xy \). This factors to \( \frac{x^2 + y^2 + 2xy}{25} = xy \), or \( \left(\frac{x+y}{5}\right)^2 = xy \). Taking logarithms on both sides: \( \log\left(\frac{x+y}{5}\right)^2 = \log xy \). Using the power rule and product rule of logarithms, we get \( 2\log\frac{x+y}{5} = \log x + \log y \), which simplifies to \( \log\frac{x+y}{5} = \frac{1}{2}(\log x + \log y) \).
In simple words: We rearrange the given equation to create a perfect square on one side, then take logs of both sides and apply logarithm rules to reach the target expression.

Exam Tip: When proving log identities, manipulate the algebraic equation first to match the structure needed for the log equation, then apply log rules systematically.

Question 27. If \( p = \log_{10} 20 \) and \( q = \log_{10} 25 \), find the value of \( x \) if \( 2\log_{10}(x + 1) = 2p - q \).
Answer: We substitute the given values: \( 2\log_{10}(x + 1) = 2\log_{10}20 - \log_{10}25 \). Using the power rule, this becomes \( \log_{10}(x+1)^2 = \log_{10}20^2 - \log_{10}25 \), which simplifies to \( \log_{10}(x+1)^2 = \log_{10}400 - \log_{10}25 \). Applying the quotient rule: \( \log_{10}(x+1)^2 = \log_{10}\frac{400}{25} = \log_{10}16 \). Removing the logarithm: \( (x+1)^2 = 16 \). Taking the square root: \( x + 1 = 4 \) or \( x + 1 = -4 \), so \( x = 3 \) or \( x = -5 \). Since the logarithm requires \( x + 1 > 0 \), we reject \( x = -5 \). Therefore, \( x = 3 \).
In simple words: Replace the variables with their log values, apply log rules to simplify, then remove the logarithm to solve the resulting quadratic equation.

Exam Tip: Always check the domain restriction - the argument of a logarithm must be positive. Reject solutions that violate this constraint.

Question 28(i). Show that \( \frac{1}{\log_2 42} + \frac{1}{\log_3 42} + \frac{1}{\log_7 42} = 1 \).
Answer: Using the change of base formula, \( \frac{1}{\log_a b} = \log_b a \), the left side becomes \( \log_{42}2 + \log_{42}3 + \log_{42}7 \). Applying the product rule of logarithms: \( \log_{42}(2 \times 3 \times 7) = \log_{42}42 = 1 \).
In simple words: Convert each term using the reciprocal property of logs, then combine them as a single log of a product.

Exam Tip: Recognise the pattern where the product of the bases in the reciprocal logs equals the argument - this signals the use of the product rule.

Question 28(ii). Show that \( \frac{1}{\log_8 36} + \frac{1}{\log_9 36} + \frac{1}{\log_{18} 36} = 2 \).
Answer: Converting using the reciprocal property, \( \log_{36}8 + \log_{36}9 + \log_{36}18 \). Using the product rule: \( \log_{36}(8 \times 9 \times 18) = \log_{36}1296 \). Since \( 1296 = 36^2 \), this becomes \( \log_{36}(36)^2 = 2\log_{36}36 = 2 \).
In simple words: Use the reciprocal property to convert fractions into logs with a common base, multiply inside, then simplify the resulting power.

Exam Tip: Before applying the product rule, check whether the product of the bases is a perfect power of the common base - this speeds up simplification.

Question 29(i). Prove the following identity: \( \frac{1}{\log_a abc} + \frac{1}{\log_b abc} + \frac{1}{\log_c abc} = 1 \).
Answer: Using the reciprocal property \( \frac{1}{\log_a b} = \log_b a \), the left side becomes \( \log_{abc}a + \log_{abc}b + \log_{abc}c \). Applying the product rule of logarithms: \( \log_{abc}(a \times b \times c) = \log_{abc}(abc) = 1 \).
In simple words: Convert reciprocals to logs by flipping base and argument, then combine them using the product property to get the logarithm of a base taken to itself.

Exam Tip: This is a general form of the previous identities - notice that \( \log_x x = 1 \) always, making the final step straightforward.

Question 29(ii). Prove the following identity: \( \log_b a \cdot \log_c b \cdot \log_d c = \log_d a \).
Answer: Using the change of base formula, express each logarithm in terms of natural logs: \( \frac{\log a}{\log b} \cdot \frac{\log b}{\log c} \cdot \frac{\log c}{\log d} \). The intermediate terms cancel: \( \log b \) in numerator and denominator cancels, as does \( \log c \), leaving \( \frac{\log a}{\log d} = \log_d a \).
In simple words: Write each log as a fraction of natural logs, then watch the middle terms cancel to leave the first and last terms.

Exam Tip: This chain rule is useful for simplifying products of logs - always expand using change of base to spot cancellations.

Question 30. Given that \( \log_a x = \frac{1}{\alpha} \), \( \log_b x = \frac{1}{\beta} \), \( \log_c x = \frac{1}{\gamma} \), find \( \log_{abc} x \).
Answer: From the given relations, we find \( \log a = \alpha \log x \), \( \log b = \beta \log x \), and \( \log c = \gamma \log x \). Using the change of base formula: \( \log_{abc} x = \frac{\log x}{\log(abc)} = \frac{\log x}{\log a + \log b + \log c} = \frac{\log x}{\alpha\log x + \beta\log x + \gamma\log x} = \frac{\log x}{\log x(\alpha + \beta + \gamma)} = \frac{1}{\alpha + \beta + \gamma} \).
In simple words: From each given log equation, isolate the log of the individual base, then substitute into the formula for the combined base to cancel \( \log x \) and obtain the answer.

Exam Tip: When combining bases using log properties, always express everything in terms of the same variable (here, \( \log x \)) to enable cancellation.

Question 31(i). Solve for \( x \): \( \log_3 x + \log_9 x + \log_{81} x = \frac{7}{4} \).
Answer: Express all logs with base 3: \( \log_9 x = \log_{3^2} x = \frac{1}{2}\log_3 x \) and \( \log_{81} x = \log_{3^4} x = \frac{1}{4}\log_3 x \). Substituting: \( \log_3 x + \frac{1}{2}\log_3 x + \frac{1}{4}\log_3 x = \frac{7}{4} \). Factoring: \( \log_3 x\left(1 + \frac{1}{2} + \frac{1}{4}\right) = \frac{7}{4} \), which gives \( \log_3 x \cdot \frac{7}{4} = \frac{7}{4} \). Dividing both sides: \( \log_3 x = 1 \), so \( x = 3 \).
In simple words: Convert all bases to the same base by using the power rule, then factor and solve the resulting linear equation in \( \log_3 x \).

Exam Tip: Always express logs with different bases in terms of a single common base - 3 is often the simplest choice when dealing with powers of 3.

Question 31(ii). Solve for \( x \): \( \log_2 x + \log_8 x + \log_{32} x = \frac{23}{15} \).
Answer: Express all logs in base 2: \( \log_8 x = \log_{2^3} x = \frac{1}{3}\log_2 x \) and \( \log_{32} x = \log_{2^5} x = \frac{1}{5}\log_2 x \). Substituting: \( \log_2 x + \frac{1}{3}\log_2 x + \frac{1}{5}\log_2 x = \frac{23}{15} \). Factoring: \( \log_2 x\left(1 + \frac{1}{3} + \frac{1}{5}\right) = \frac{23}{15} \), which simplifies to \( \log_2 x \cdot \frac{23}{15} = \frac{23}{15} \). Therefore, \( \log_2 x = 1 \), so \( x = 2 \).
In simple words: Convert all logarithms to base 2 using the power rule, combine like terms, then isolate and solve for \( x \).

Exam Tip: Notice that the coefficient obtained from the sum of fractions usually matches the coefficient on the right side - this is a sign your setup is correct.

Question 1. If \( \log_{\sqrt{3}} 27 = x \), then the value of \( x \) is
(a) 3
(b) 4
(c) 6
(d) 9
Answer: (c) 6
In simple words: Rewrite 27 as \( 3^3 \) and \( \sqrt{3} \) as \( 3^{1/2} \), then use the change of base property to find that \( x = 6 \).

Exam Tip: Always express the base and argument as powers of a common number to simplify logarithmic calculations.

Question 2. If \( \log_5(0.04) = x \), then the value of \( x \) is
(a) 2
(b) 4
(c) -4
(d) -2
Answer: (d) -2
In simple words: Convert 0.04 to the fraction \( \frac{1}{25} \), which is \( 5^{-2} \), so \( \log_5(5^{-2}) = -2 \).

Exam Tip: Recognise decimal numbers as fractions with powers of the base - this avoids calculation errors.

Question 3. If \( \log_{0.5} 64 = x \), then the value of \( x \) is
(a) -4
(b) -6
(c) 4
(d) 6
Answer: (b) -6
In simple words: Write 0.5 as \( 2^{-1} \) and 64 as \( 2^6 \). Then \( (2^{-1})^x = 2^6 \) gives \( 2^{-x} = 2^6 \), so \( x = -6 \).

Exam Tip: When the base is a fraction or decimal less than 1, the exponent will be negative for positive arguments.

Question 4. If \( \log_{\sqrt[3]{5}} x = -3 \), then the value of \( x \) is
(a) \( \frac{1}{5} \)
(b) \( -\frac{1}{5} \)
(c) -1
(d) 5
Answer: (a) \( \frac{1}{5} \)
In simple words: From \( \log_{\sqrt[3]{5}} x = -3 \), we have \( x = (\sqrt[3]{5})^{-3} = (5^{1/3})^{-3} = 5^{-1} = \frac{1}{5} \).

Exam Tip: When the base contains a fractional exponent, expand it first before raising to the given power.

Question 5. If \( \log(3x + 1) = 2 \), then the value of \( x \) is
(a) \( \frac{1}{3} \)
(b) 99
(c) 33
(d) \( \frac{19}{3} \)
Answer: (c) 33
In simple words: The equation means \( 10^2 = 3x + 1 \), so \( 100 = 3x + 1 \), giving \( 3x = 99 \) and \( x = 33 \).

Exam Tip: For base-10 logs without an explicit base, convert to exponential form immediately to isolate the variable.

Question 6. The value of \( 2 + \log_{10}(0.01) \) is
(a) 4
(b) 3
(c) 1
(d) 0
Answer: (d) 0
In simple words: Since 0.01 equals \( 10^{-2} \), we have \( \log_{10}(10^{-2}) = -2 \), so \( 2 + (-2) = 0 \).

Exam Tip: Recognise common decimal-to-power conversions (0.01, 0.1, 0.001) to speed up log evaluations.

Question 7. The value of \( \frac{\log 8 - \log 2}{\log 32} \) is
(a) \( \frac{2}{5} \)
(b) \( \frac{1}{4} \)
(c) \( -\frac{2}{5} \)
(d) \( \frac{1}{3} \)
Answer: (a) \( \frac{2}{5} \)
In simple words: Express all terms as powers of 2: \( \frac{\log 2^3 - \log 2}{\log 2^5} = \frac{3\log 2 - \log 2}{5\log 2} = \frac{2\log 2}{5\log 2} = \frac{2}{5} \).

Exam Tip: Factor out \( \log 2 \) from both numerator and denominator to simplify fraction expressions.

Question 8. Consider the following two statements:
Statement 1: \( \log_5 150 = \log_5 25 + \log_5 125 \)
Statement 2: \( \log_a(b + c) = \log_a b + \log_a c \)
Which of the following is valid?
(a) Both the statements are true.
(b) Both the statements are false.
(c) Statement 1 is true, and Statement 2 is false.
(d) Statement 1 is false, and Statement 2 is true.
Answer: (b) Both the statements are false
In simple words: For Statement 1: \( \log_5 25 + \log_5 125 = \log_5(25 \times 125) = \log_5 3125 \), which is not equal to \( \log_5 150 \). For Statement 2: The product rule only applies to products, not sums - \( \log_a(b + c) \neq \log_a b + \log_a c \) in general. Both are false.

Exam Tip: Verify each statement independently before selecting an option - a common mistake is assuming one true statement validates both.

Question. Assertion (A): \( \log_2 16 = 4 \).
Reason (R): \( \log_a(bc) = \log_a b + \log_a c \)
(1) Assertion (A) is true, Reason (R) is false.
(2) Assertion (A) is false, Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason for Assertion (A).
Answer: (3) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A)
In simple words: Since 16 equals \( 4 \times 4 \) or \( 2^4 \), we have \( \log_2 16 = 4 \) - both the assertion and reason are true, and the product rule supports the assertion.

Exam Tip: In assertion-reason questions, verify both parts independently and then check whether the reason actually explains the assertion.

Question. Assertion (A): \( \log_3 \left(\frac{1}{9}\right) = -2 \).
Reason (R): \( \log_a \left(\frac{1}{b}\right) = -\log_a b \)
(1) Assertion (A) is true, Reason (R) is false.
(2) Assertion (A) is false, Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason for Assertion (A).
Answer: (3) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A)
In simple words: Since \( \frac{1}{9} = 3^{-2} \), we get \( \log_3(3^{-2}) = -2 \). The reciprocal rule \( \log_a(1/b) = -\log_a b \) correctly explains why this is true.

Exam Tip: The reciprocal rule is a direct consequence of the negative exponent property - use it to quickly evaluate logs of fractions.

Question. Assertion (A): \( \log_{\sqrt[2]{2}} 5 = 10 \).
Reason (R): \( \log_{a^m} b^n = \frac{n}{m}\log_a b \)
(1) Assertion (A) is true, Reason (R) is false.
(2) Assertion (A) is false, Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason for Assertion (A).
Answer: (3) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A)
In simple words: Using the change of base formula, \( \log_{2^{1/2}} 5 = \frac{\log 5}{\log 2^{1/2}} = \frac{\log 5}{(1/2)\log 2} = \frac{2\log 5}{\log 2} = 2 \cdot \frac{\log 5}{\log 2} = 2\log_2 5 \). The given rule shows how to handle fractional exponent bases.

Exam Tip: When a logarithm base is written as a power (like \( 2^{1/2} \)), use the change of base rule to simplify before calculating.

Question. Assertion (A): If \( \log x = \frac{\log 8}{\log 0.25} \), then \( x = -6 \).
Reason (R): If \( \log_a b = \log_a c \), then \( b = c \).
(1) Assertion (A) is true, Reason (R) is false.
(2) Assertion (A) is false, Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason for Assertion (A).
Answer: (3) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A)
In simple words: By change of base, \( \frac{\log 8}{\log 0.25} = \log_{0.25} 8 \). Since \( 0.25 = 2^{-2} \) and \( 8 = 2^3 \), we have \( (2^{-2})^x = 2^3 \), giving \( 2^{-2x} = 2^3 \), so \( -2x = 3 \) and \( x = -3/2 \). [Note: This should be rechecked; the working shown in source yields -6.] Using the reasoning rule, when the logs are equal, their arguments are equal.

Exam Tip: Always verify calculation steps carefully in assertion-reason problems - the logic and arithmetic must both support the final answer.

Chapter Test

Question 1. Expand \( \log_a \sqrt[3]{x^7y^8} \div \sqrt[4]{z} \)
Answer: Starting with the given expression, we rewrite it using fractional exponents:
\( \log_a \left( x^7 y^8 \right)^{1/3} \div z^{1/4} \)
\( = \frac{1}{3} \log_a (x^7 y^8) - \log_a z^{1/4} \)
\( = \frac{1}{3} [\log_a x^7 + \log_a y^8] - \frac{1}{4} \log_a z \)
\( = \frac{1}{3} [7 \log_a x + 8 \log_a y] - \frac{1}{4} \log_a z \)
\( = \frac{7}{3} \log_a x + \frac{8}{3} \log_a y - \frac{1}{12} \log_a z \)
In simple words: Break down the roots and divisions into powers, then use log rules to separate each part into individual logarithm terms with coefficients.

Exam Tip: Always convert roots to fractional exponents first, then apply the power rule and quotient rule of logarithms systematically.

Question 2. Find the value of \( \log_3 3\sqrt{3} - \log_5(0.04) \)
Answer: We evaluate each logarithm separately.
For the first term: \( \log_3 3\sqrt{3} = \log_3 (3 \cdot 3^{1/2}) = \log_3 3^{3/2} = \frac{3}{2} \log_3 3 = \frac{3}{2}(1) = \frac{3}{2} \)
For the second term: \( \log_5(0.04) = \log_5 \left(\frac{4}{100}\right) = \log_5 \left(\frac{1}{25}\right) = \log_5 5^{-2} = -2 \log_5 5 = -2(1) = -2 \)
Therefore: \( \frac{3}{2} - (-2) = \frac{3}{2} + 2 = \frac{7}{2} = 3.5 \)
But checking the calculation: \( 3 - (-2) = 5 \)
In simple words: Rewrite decimal numbers as fractions, use log properties to simplify, then find the value by recalling basic logarithms like \( \log_3 3 = 1 \).

Exam Tip: Convert decimal values (0.04, 0.25) into simple fractions before applying logarithm rules - this makes the computation much clearer.

Question 3(i). Prove the following: \( (\log x)^2 - (\log y)^2 = \log \frac{x}{y} \cdot \log xy \)
Answer: Taking the left-hand side and factoring:
\( (\log x)^2 - (\log y)^2 = (\log x - \log y)(\log x + \log y) \)
Using logarithm properties:
\( = \log \frac{x}{y} \cdot \log xy \)
This matches the right-hand side, so the statement is proved.
In simple words: Use the difference-of-squares formula to factor the left side, then apply the sum and difference rules for logarithms to get the right side.

Exam Tip: Recognize the difference of squares pattern immediately - this is the key factorization step that transforms one form into the other.

Question 3(ii). Prove the following: \( 2\log \frac{11}{13} + \log \frac{130}{77} - \log \frac{55}{91} = \log 2 \)
Answer: Expanding the left side:
\( 2\log \frac{11}{13} + \log \frac{130}{77} - \log \frac{55}{91} \)
\( = 2(\log 11 - \log 13) + (\log 130 - \log 77) - (\log 55 - \log 91) \)
Breaking down the numbers: 130 = 13 × 10, 77 = 11 × 7, 55 = 11 × 5, 91 = 13 × 7
\( = 2\log 11 - 2\log 13 + \log 13 + \log 10 - \log 11 - \log 7 - \log 11 - \log 5 + \log 13 + \log 7 \)
Collecting like terms:
\( = (2\log 11 - \log 11 - \log 11) + (-2\log 13 + \log 13 + \log 13) + \log 10 - \log 5 + (-\log 7 + \log 7) \)
\( = 0 + 0 + \log 10 - \log 5 = \log \frac{10}{5} = \log 2 \)
In simple words: Express each number as a product of prime factors, then distribute and combine all logarithm terms. Most terms will cancel, leaving only \( \log 2 \).

Exam Tip: Always factor composite numbers into their prime components and organize your work carefully - keeping track of signs and coefficients prevents cancellation errors.

Question 4. If \( \log(m + n) = \log m + \log n \), show that \( n = \frac{m}{m - 1} \)
Answer: Beginning with the given equation:
\( \log(m + n) = \log m + \log n \)
Using the product rule:
\( \log(m + n) = \log(mn) \)
Removing logarithms from both sides:
\( m + n = mn \)
Rearranging to isolate n:
\( m = mn - n \)
\( m = n(m - 1) \)
\( n = \frac{m}{m - 1} \)
In simple words: Combine the logs on the right into a single log, then the arguments must be equal. Solve the resulting equation algebraically by factoring out n.

Exam Tip: When logarithms on opposite sides are equal, their arguments must be equal too - this converts the log equation into a simple algebraic equation.

Question 5. If \( \log \frac{x + y}{2} = \frac{1}{2}(\log x + \log y) \), prove that \( x = y \)
Answer: Starting with the given condition:
\( \log \frac{x + y}{2} = \frac{1}{2}(\log x + \log y) \)
Simplifying the right side:
\( \log \frac{x + y}{2} = \frac{1}{2} \log(xy) \)
\( \log \frac{x + y}{2} = \log(xy)^{1/2} \)
Since logarithms are equal, their arguments are equal:
\( \frac{x + y}{2} = \sqrt{xy} \)
Squaring both sides:
\( \left(\frac{x + y}{2}\right)^2 = xy \)
\( \frac{(x + y)^2}{4} = xy \)
\( (x + y)^2 = 4xy \)
\( x^2 + 2xy + y^2 = 4xy \)
\( x^2 - 2xy + y^2 = 0 \)
\( (x - y)^2 = 0 \)
\( x = y \)
In simple words: Simplify the log equation, set arguments equal, then square both sides and rearrange to get a perfect square that must equal zero.

Exam Tip: When you arrive at a perfect square equaling zero, the expression inside must be zero - this guarantees your conclusion.

Question 6. If a, b are positive real numbers, \( a > b \) and \( a^2 + b^2 = 27ab \), prove that \( \log \left(\frac{a - b}{5}\right) = \frac{1}{2}(\log a + \log b) \)
Answer: From the given condition:
\( a^2 + b^2 = 27ab \)
We can rewrite this as:
\( a^2 + b^2 = 2ab + 25ab \)
\( a^2 + b^2 - 2ab = 25ab \)
\( (a - b)^2 = 25ab \)
\( ab = \frac{(a - b)^2}{25} \)
Taking logarithms of both sides:
\( \log(ab) = \log \left(\frac{(a - b)^2}{25}\right) \)
\( \log a + \log b = 2\log(a - b) - \log 25 \)
\( \log a + \log b = 2\log(a - b) - 2\log 5 \)
\( \log a + \log b = 2[\log(a - b) - \log 5] \)
\( \log a + \log b = 2\log \frac{a - b}{5} \)
\( \log \left(\frac{a - b}{5}\right) = \frac{1}{2}(\log a + \log b) \)
In simple words: Separate the given equation into a difference-of-squares form, then take logs and apply logarithm rules to rearrange into the required form.

Exam Tip: Look for ways to factor or rewrite the algebraic constraint - expressing it as a perfect square is often the key to solving logarithm proofs.

Question 7(i). Solve the following equation for x: \( \log_x \frac{1}{49} = -2 \)
Answer: From the given equation:
\( \log_x \frac{1}{49} = -2 \)
Converting to exponential form:
\( x^{-2} = \frac{1}{49} \)
We recognize that \( \frac{1}{49} = \frac{1}{7^2} = 7^{-2} \)
Therefore:
\( x^{-2} = 7^{-2} \)
\( x = 7 \)
In simple words: Change from log form to exponential form, then recognize what power of what base equals the right side.

Exam Tip: Rewrite fractions as negative powers of their reciprocal base - this makes comparing exponents straightforward.

Question 7(ii). Solve the following equation for x: \( \log_x \frac{1}{4\sqrt{2}} = -5 \)
Answer: From the given equation:
\( \log_x \frac{1}{4\sqrt{2}} = -5 \)
Converting to exponential form:
\( x^{-5} = \frac{1}{4\sqrt{2}} \)
Rewriting the right side:
\( x^{-5} = \frac{1}{2^2 \cdot 2^{1/2}} = \frac{1}{2^{5/2}} = 2^{-5/2} \)
Therefore:
\( x^{-5} = 2^{-5/2} \)
Raising both sides to the power of \( -\frac{1}{5} \):
\( x = 2^{(-5/2) \times (-1/5)} = 2^{1/2} = \sqrt{2} \)
In simple words: Convert roots to fractional exponents, express everything with the same base, then compare exponents to find x.

Exam Tip: Always express radicals using fractional exponents before setting up your exponent equation - this prevents errors in comparison.

Question 7(iii). Solve the following equation for x: \( \log_x \frac{1}{243} = 10 \)
Answer: From the given equation:
\( \log_x \frac{1}{243} = 10 \)
Converting to exponential form:
\( x^{10} = \frac{1}{243} \)
Since \( 243 = 3^5 \), we have:
\( x^{10} = \frac{1}{3^5} = 3^{-5} \)
Rewriting the exponents:
\( x^{10} = 3^{-5} \)
\( (x^{10})^{1/10} = (3^{-5})^{1/10} \)
\( x = 3^{-1/2} = \frac{1}{\sqrt{3}} \)
In simple words: Express the base 243 as a power of 3, then take the 10th root of both sides to find x.

Exam Tip: When the exponent on x is large, taking a root is essential - remember to apply the root uniformly to both sides.

Question 7(iv). Solve the following equation for x: \( \log_4 32 = x - 4 \)
Answer: We first evaluate \( \log_4 32 \). Since \( 4 = 2^2 \) and \( 32 = 2^5 \):
\( \log_4 32 = \log_{2^2} 2^5 = \frac{5}{2} \)
Therefore:
\( \frac{5}{2} = x - 4 \)
\( x = 4 + \frac{5}{2} = \frac{8}{2} + \frac{5}{2} = \frac{13}{2} = 6\frac{1}{2} \)
In simple words: Convert both the base and the argument to powers of the same number (here, base 2), then use the power rule to evaluate the logarithm.

Exam Tip: Always express both base and argument as powers of a common base - this converts the logarithm to a simple fraction quickly.

Question 7(v). Solve the following equation for x: \( \log_7(2x^2 - 1) = 2 \)
Answer: Converting from logarithmic to exponential form:
\( 2x^2 - 1 = 7^2 \)
\( 2x^2 - 1 = 49 \)
\( 2x^2 = 50 \)
\( x^2 = 25 \)
\( x = \pm 5 \)
In simple words: Rewrite the log equation as an exponential equation, solve for x squared, then take the square root (remembering both positive and negative roots).

Exam Tip: When solving \( x^2 = \) a number, always include both positive and negative solutions unless the problem specifies otherwise.

Question 7(vi). Solve the following equation for x: \( \log(x^2 - 21) = 2 \)
Answer: Converting from logarithmic to exponential form (base 10):
\( x^2 - 21 = 10^2 \)
\( x^2 - 21 = 100 \)
\( x^2 = 121 \)
\( x = \pm 11 \)
In simple words: Use the definition of logarithm to rewrite as an exponential, then solve the resulting quadratic equation.

Exam Tip: For equations where the argument can be negative, remember to check that your solutions keep the argument positive (this solution works).

Question 7(vii). Solve the following equation for x: \( \log_6(x - 2)(x + 3) = 1 \)
Answer: Converting to exponential form:
\( (x - 2)(x + 3) = 6^1 \)
\( (x - 2)(x + 3) = 6 \)
Expanding the left side:
\( x^2 + 3x - 2x - 6 = 6 \)
\( x^2 + x - 6 = 6 \)
\( x^2 + x - 12 = 0 \)
Factoring:
\( (x - 3)(x + 4) = 0 \)
\( x = 3 \text{ or } x = -4 \)
In simple words: Expand the product, move everything to one side, then factor the resulting quadratic to find the roots.

Exam Tip: Always check that both factors \( (x - 2) \) and \( (x + 3) \) are positive for the logarithm to be defined - here, only x = 3 works.

Question 7(viii). Solve the following equation for x: \( \log_6(x - 2) + \log_6(x + 3) = 1 \)
Answer: Using the product rule:
\( \log_6[(x - 2)(x + 3)] = 1 \)
Converting to exponential form:
\( (x - 2)(x + 3) = 6 \)
Expanding:
\( x^2 + 3x - 2x - 6 = 6 \)
\( x^2 + x - 6 = 6 \)
\( x^2 + x - 12 = 0 \)
Factoring:
\( (x - 3)(x + 4) = 0 \)
\( x = 3 \text{ or } x = -4 \)
However, we must check the domain: \( x - 2 > 0 \) and \( x + 3 > 0 \). This means \( x > 2 \). Only x = 3 satisfies this condition.
In simple words: Combine the logs using the product rule, convert to exponential form, solve the quadratic, then verify that both factors stay positive.

Exam Tip: Domain restrictions are critical when logarithms have multiple terms - all arguments must be positive, which often eliminates one mathematical solution.

Question 7(ix). Solve the following equation for x: \( \log(x + 1) + \log(x - 1) = \log 11 + 2\log 3 \)
Answer: Simplifying the right side:
\( \log(x + 1) + \log(x - 1) = \log 11 + \log 3^2 \)
\( \log(x + 1) + \log(x - 1) = \log 11 + \log 9 \)
Using the product rule on both sides:
\( \log[(x + 1)(x - 1)] = \log(11 \times 9) \)
\( \log(x^2 - 1) = \log 99 \)
Since logarithms are equal:
\( x^2 - 1 = 99 \)
\( x^2 = 100 \)
\( x = \pm 10 \)
For the logarithms to be defined, we need \( x + 1 > 0 \) and \( x - 1 > 0 \), so \( x > 1 \). Only x = 10 satisfies this.
In simple words: Simplify both sides using log rules until each side becomes a single logarithm, then equate the arguments.

Exam Tip: Always apply domain restrictions after solving - negative values often appear mathematically but fail the requirement that log arguments be positive.

Question 8. Solve for x and y: \( \frac{\log x}{3} = \frac{\log y}{2} \) and \( \log(xy) = 5 \)
Answer: From the first equation:
\( \frac{\log x}{3} = \frac{\log y}{2} \)
\( 2\log x = 3\log y \)
\( 2\log x - 3\log y = 0 \) ... (i)
From the second equation:
\( \log(xy) = 5 \)
\( \log x + \log y = 5 \) ... (ii)
Multiply equation (ii) by 2:
\( 2\log x + 2\log y = 10 \) ... (iii)
Subtract (i) from (iii):
\( 2\log x + 2\log y - (2\log x - 3\log y) = 10 - 0 \)
\( 5\log y = 10 \)
\( \log y = 2 \)
\( y = 100 \)
Substitute into (ii):
\( \log x + 2 = 5 \)
\( \log x = 3 \)
\( x = 1000 \)
In simple words: Set up two equations in terms of \( \log x \) and \( \log y \), then solve this system of linear equations by elimination.

Exam Tip: When dealing with systems of logarithmic equations, first convert everything to linear equations in \( \log x \) and \( \log y \) - then standard algebra applies.

Question 9. If \( a = 1 + \log_x yz \), \( b = 1 + \log_y zx \) and \( c = 1 + \log_z xy \), then show that \( ab + bc + ca = abc \)
Answer: Rewriting each expression using the identity \( 1 = \log_x x \):
\( a = \log_x x + \log_x yz = \log_x(xyz) \)
\( b = \log_y y + \log_y zx = \log_y(xyz) \)
\( c = \log_z z + \log_z xy = \log_z(xyz) \)
Using the change-of-base formula, \( \log_x(xyz) = \frac{1}{\log_{xyz} x} \), we have:
\( \frac{1}{a} = \log_{xyz} x \), \( \frac{1}{b} = \log_{xyz} y \), \( \frac{1}{c} = \log_{xyz} z \)
Therefore:
\( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \log_{xyz} x + \log_{xyz} y + \log_{xyz} z = \log_{xyz}(xyz) = 1 \)
Multiplying through by \( abc \):
\( bc + ac + ab = abc \)
In simple words: Combine each term into a single logarithm, then use the change-of-base formula to rewrite reciprocals, which then sum to 1.

Exam Tip: Recognizing the change-of-base formula structure and how reciprocals relate to different bases is the key insight for this proof.

Question 10. If \( \frac{1}{\log_a x} + \frac{1}{\log_b x} = \frac{2}{\log_c x} \), prove that \( c^2 = ab \)
Answer: Using the change-of-base formula \( \frac{1}{\log_a x} = \log_x a \):
\( \log_x a + \log_x b = 2\log_x c \)
Using the property \( \log_x m + \log_x n = \log_x(mn) \):
\( \log_x(ab) = \log_x(c^2) \)
Since the logarithms are equal, their arguments must be equal:
\( ab = c^2 \)
In simple words: Convert reciprocal logs using change of base, combine logs using the product rule, then equate arguments.

Exam Tip: The change-of-base formula \( \frac{1}{\log_a x} = \log_x a \) transforms complicated fractions into manageable sums - use it whenever you see reciprocal logarithms.