Exercise 2.1
Question 1. Simplify the following:-
(i) 3(a4b3)10 × 5(a2b2)3
(ii) (2x -2 y3)3
(iii) ((4 × 107 ) (6 × 10-5))/(8 × 104 )
(iv) (4ab2 (-5ab3 ))/(10a2 b2 )
(v) {(x2 y2)/(a2 b2 )}n
(vi) (a3n-9)6/a2n-4
Solution 1.
(i) We have 3(a4b3)10 × 5(a2b2)3
By using identity:- (xm)n = xmn
After simplification we get
= 3(a40b30) × 5(a6b6)
= 3 × a40 × b30 × 5 × a6 × b6
By using identity:- xm × xn = xm+n
= 15 × a46 × b36
= 15a46 b36 Answer.
(ii) (2x -2 y3)3
By using identity:- (x-m)= 1/xm and (xm)n = xmn
After simplification we get
= 23 × 1/x2x3 × y3×3
= 8 × 1/x6 × y9
= 8 y9
= 8 x-6 y6 Answer.
Question 2. If a = 3 and b = - 2, find the value of:
(i) aa + bb
(ii) ab + ba
(iii) (a+b)ab
Solution 2.
(i) We have aa + bb
Put the value a = 3 and b = -2 in equation given above
= 33 + (-2)-2
= 3 × 3 × 3 + 1/((-22))
= 27 +1/((-2 ×-2))
= 27 +1/4
= 27/1+1/4
Take the LCM of denominators 1 and 4 = 4
= (108+1)/4
= 109/4 Answer.
(ii) We have ab + ba
Put the value a = 3 and b = –2 in equation given above
= 3-2 + (–2)3
= 1/((32)) + (–2 × –2 × –2)
= 1/9 – 8
= 1/9-8/1
Take the LCM of denominators 9 and 1 = 9
= (1-72)/9
= -71/9 Answer.
(iii) We have (a+b)ab
Put the value a = 3 and b = –2 in equation given above
= {3 + (–2)}3×(–2)
= {3 + (–2)}–6
= {3 – 2}–6
= (1) –6
= 1/16
= 1 Answer.
Question 3. Prove that:
Solution 3.
Question 4. Prove that:
Solution 4.
Question 5. Prove that:
Solution 5.
Question 8. Solve the following equation for x:
(i) 7(2x+3) = 1
(ii) 2(x+1)= 4(x-3)
(iii) 2(5x+3) = 8(x+3)
(iv) 42x = 1/32
(v) 4(x-1)×(0.5)(3-2x) = (1/8)x
(vi) 2(3x-7) = 256
Solution 8.
(i) we have 7(2x+3) = 1
By using identity:- x0 =1
7(2x+3) = 70
We should calculate the value of x through their powers.
2x + 3 = 0
2x = -3
x = (-3)/2 Answer.
(ii) we have 2(x+1)= 4(x-3)
By using identity:- (xm)n = xmn
2(x+1)= (22)(x-3)
2(x+1)= (2)2(x-3)
2(x+1) = (2)(2x-6)
We should calculate the value of x through their powers.
x + 1 = 2x – 6
x – 2x = – 6 – 1
– x = – 7
x = 7 Answer.
(iii) we have 2(5x+3) = 8(x+3)
By using identity:- (xm)n = xmn
2(5x+3) = (23)(x+3)
2(5x+3)=(2)3(x+3)
2(5x+3)=2(3x+9)
We should calculate the value of x through their powers.
5x+3 = 3x+9
5x – 3x = 9 – 3
2x = 6
x = 6/2
x = 3 Answer.
(iv) we have 42x = 1/32
(4)2x = 1/(2)5
(22)2x = 1/(2)5
By using identity:- (xm)n = xmn
(2)4x = 2(-5)
We should calculate the value of x through their powers.
4x = -5
x = -5/4 Answer.
(vi) 2(3x-7) = 256
We should calculate the value of x through their powers.
2(3x-7) = 28
3x - 7 = 8
3x = 8 + 7
3x = 15
x = 15/3
x = 5 Answer.
Question 9. Solve the following equations for x:
(i) 22x -22x+3+24 = 0
(ii) 3(2x+4)+1=2×3(x+2)
Solution 9.
Question 10. If 49392 = a4b2c3, find the value of a, b and c where a, b and c are different positive primes.
Solution 10.
It is Given that 49392 = a4b2c3
By prime factorisation method we get
By prime factorisation we can observe that 24,32 and 73.
Hence the value of a = 2, b = 3 and c = 7 Answer.
Question 11. If 1176 = 2a3b7c, find a,b and c.
Solution 11
It is Given that 1176 = 2a3b7c
By prime factorisation method we get
By prime factorisation we can observe that 23,31 and 72.
Hence the value of a = 3, b = 1 and c = 2 Answer.
Question 12. Given 4725 = 3a 5b 7c, find
(i) the integral value of a, b and c
(ii) the value of 2-a3b7c
Solution 12.
(i) It is Given that 4725 = 3a 5b 7c
By prime factorisation method we get
By prime factorisation we can observe that 33,52 and 71.
Hence the value of a = 3, b = 2 and c = 1.
Exercise 2.2
Question 1. Assuming that x,y,z are positive real numbers, simplify each of the following:
Solution 1.
Exercise 2.3
Question 1. Write (625)-1/4 in decimal form.
Solution 1.
We have (625)-1/4
By the prime factorisation 625 we get 5×5×5×5
= (54)-1/4
= 5-4/4
= 5-1
= 1/5
= 0.2
Question 2. State the product law of exponents.
Solution 2.
The product law of exponents is, “If a is any real number and m, n are positive integers, then xm × xn = x(m+n)
With the reference of above definition we get,
xm × xn =(x×x×x……………to m factors)×(x×x×x……………to n factors)
= a×a×a……………….. to (m+n) factors
= a(m+n)
Hence, am×an = a(m+n)
Question 3. State the quotient law of exponents.
Solution 3.
In the quotient law of exponents we need to divide two powers with the same base as given in example:- xm/xn
Here a≠0.
When base is same and exponents in division we should subtracting the exponents. Here m and n are the positive integers, then xm/xn =x(m-n)
Question 4. State the power law of exponents.
Solution 4.
In the power rule of exponents we need to raise a power to a power by multiply the exponents with each other. If a is any real number and m, n are positive integers, then
each other. If a is any real number and m, n are positive integers, then
(xm )n =xm× xm× xm × xm……………..n factors
(xm )n =(x×x×x×…..m)× (x×x×x×…..n)
(xm )n =(x×x×x…..mn) factors
(xm )n = xmn
Question 5. If 24×42=16x, then find the value of x.
Solution 5.
We have 24×42 = 16x
By the prime factorisation of 16 we get 2×2×2×2.
24×42 = (24 )x
24×(22)2=(24)x
24 × 24=(24)x
28=24x
By equating the exponents we get
8 = 4x
8/4=x
2=x Answer
Question 6. If 3x-1=9 and 4y+2=64, what is the value of x/y?
Solution 6.
We have 3x-1=9
By the prime factorisation of 9 we get 3×3
3x-1 = 3×3
3x-1 = 32
By equating the exponents we get
x-1=2
x=2+1
x=3
4y+2 = 64
By the prime factorisation of 64 we get 4×4×4
4y+2 = 4×4×4
4y+2 = 43
By equating the exponents we get
y+2=3
y=3-2
y=1
Therefore,
x/y=3/1
x/y=3 Answer
Question 14. If (x-1)3= 8, What is the value of (x+1)2 ?
Solution 14.
We have (x-1)3 = 8
By the prime factorisation of 8 we get 2×2×2
(x-1)3 = 23
Equalising the base we get
x-1=2
x=2+1
x=3
Value of x put in given equation (x+1)2
=(3+1)2
=(4)2
=4×4
=16 Answer