Rd Sharma 2021 for Class 9 Maths Chapter 2 - Exponents Of Real Numbers

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Page No 2.12:

Question 1:

Simplify the following:

(i) $3 {(a^{4} b^{3})}^{10} \times 5 {(a^{2} b^{2})}^{3}$

(ii) ${(2 x^{- 2} y^{3})}^{3}$

(iii) $\frac{(4 \times 10^{7}) (6 \times 10^{- 5})}{8 \times 10^{4}}$

(iv) $\frac{4 a b^{2} (- 5 a b^{3})}{10 a^{2} b^{2}}$

(v) ${(\frac{x^{2} y^{2}}{a^{2} b^{3}})}^{n}$

(vi) $\frac{{(a^{3 n - 9})}^{6}}{a^{2 n - 4}}$

Answer:

(i)
$3 {(a^{4} b^{3})}^{10} \times 5 {(a^{2} b^{2})}^{3} = 3 \times a^{40} \times b^{30} \times 5 \times a^{6} \times b^{6} = 15 \times a^{40} \times a^{6} \times b^{30} \times b^{6} = 15 \times a^{40 + 6} \times b^{30 + 6} [a^{m} \times a^{n} = a^{m + n}] = 15 a^{46} b^{36}$

(ii)
${(2 x^{- 2} y^{3})}^{3} = 2^{3} \times {(x^{- 2})}^{3} \times {(y^{3})}^{3} = 8 \times x^{- 6} \times y^{9} = 8 x^{- 6} y^{9}$

(iii)
$\frac{(4 \times 10^{7}) (6 \times 10^{- 5})}{8 \times 10^{4}} = \frac{4 \times 10^{7} \times 6 \times 10^{- 5}}{8 \times 10^{4}} = \frac{24 \times 10^{7 + (- 5)}}{8 \times 10^{4}} = \frac{24 \times 10^{2}}{8 \times 10^{4}}$
$= \frac{24 \times 10^{2} \times 10^{- 4}}{8} = 3 \times 10^{2 + (- 4)} = 3 \times 10^{- 2} = \frac{3}{100}$

(iv)
$\frac{4 a b^{2} (- 5 a b^{3})}{10 a^{2} b^{2}} = \frac{4 \times a \times b^{2} \times (- 5) \times a \times b^{3}}{10 a^{2} b^{2}} = \frac{- 20 \times a^{1} \times a^{1} \times b^{2} \times b^{3}}{10 a^{2} b^{2}}$
$= \frac{- 20 \times a^{1 + 1} \times b^{2 + 3}}{10 a^{2} b^{2}} = - 2 \times a^{2} \times b^{5} \times a^{- 2} \times b^{- 2} = - 2 \times a^{2 + (- 2)} \times b^{5 + (- 2)} = - 2 \times a^{0} \times b^{3} = - 2 b^{3}$

(v)
${(\frac{x^{2} y^{2}}{a^{2} b^{3}})}^{n} = \frac{{(x^{2})}^{n} {(y^{2})}^{n}}{{(a^{2})}^{n} {(b^{3})}^{n}} = \frac{x^{2 n} y^{2 n}}{a^{2 n} b^{3 n}}$

(vi)
$\frac{{(a^{3 n - 9})}^{6}}{a^{2 n - 4}} = \frac{a^{6 (3 n - 9)}}{a^{2 n - 4}} = \frac{a^{(18 n - 54)}}{a^{2 n - 4}}$
$= a^{(18 n - 54)} \times a^{- (2 n - 4)} = a^{18 n - 54} \times a^{- 2 n + 4} = a^{18 n - 54 - 2 n + 4} = a^{16 n - 50}$

Page No 2.12:

Question 2:

If $a = 3$ and $b = - 2$ , find the values of:
(i) $a^{a} + b^{b}$
(ii) $a^{b} + b^{a}$
(iii) ${(a + b)}^{a b}$

Answer:

(i) $a^{a} + b^{b}$
Here, $a = 3$ and $b = - 2$ .
Put the values in the expression $a^{a} + b^{b}$ .
$3^{3} + {(- 2)}^{- 2} = 27 + \frac{1}{{(- 2)}^{2}} = 27 + \frac{1}{4} = \frac{108 + 1}{4} = \frac{109}{4}$

(ii) $a^{b} + b^{a}$
Here, $a = 3$ and $b = - 2$ .
Put the values in the expression $a^{b} + b^{a}$ .
$3^{- 2} + {(- 2)}^{3} = {(\frac{1}{3})}^{2} + (- 8) = \frac{1}{9} - 8 = \frac{1 - 72}{9} = - \frac{71}{9}$

(iii) ${(a + b)}^{a b}$
Here, $a = 3$ and $b = - 2$ .
Put the values in the expression ${(a + b)}^{a b}$ .
${[3 + (- 2)]}^{3 (- 2)} = {(1)}^{- 6} = 1$

Page No 2.12:

Question 3:

Prove that:

(i) ${(\frac{x^{a}}{x^{b}})}^{a^{2} + a b + b^{2}} \times {(\frac{x^{b}}{x^{c}})}^{b^{2} + b c + c^{2}} \times {(\frac{x^{c}}{x^{a}})}^{c^{2} + c a + a^{2}} = 1$
(ii) ${(\frac{x^{a}}{x^{b}})}^{c} \times {(\frac{x^{b}}{x^{c}})}^{a} \times {(\frac{x^{c}}{x^{a}})}^{b} = 1$

Answer:

(i) ${(\frac{x^{a}}{x^{b}})}^{a^{2} + a b + b^{2}} \times {(\frac{x^{b}}{x^{c}})}^{b^{2} + b c + c^{2}} \times {(\frac{x^{c}}{x^{a}})}^{c^{2} + c a + a^{2}} = 1$

Consider the left hand side:
${(\frac{x^{a}}{x^{b}})}^{a^{2} + a b + b^{2}} \times {(\frac{x^{b}}{x^{c}})}^{b^{2} + b c + c^{2}} \times {(\frac{x^{c}}{x^{a}})}^{c^{2} + c a + a^{2}} = \frac{x^{a (a^{2} + a b + b^{2})}}{x^{b (a^{2} + a b + b^{2})}} \times \frac{x^{b (b^{2} + b c + c^{2})}}{x^{c (b^{2} + b c + c^{2})}} \times \frac{x^{c (c^{2} + c a + a^{2})}}{x^{a (c^{2} + c a + a^{2})}} = x^{a (a^{2} + a b + b^{2}) - b (a^{2} + a b + b^{2})} \times x^{b (b^{2} + b c + c^{2}) - c (b^{2} + b c + c^{2})} \times x^{c (c^{2} + c a + a^{2}) - a (c^{2} + c a + a^{2})} = x^{(a - b) (a^{2} + a b + b^{2})} \times x^{(b - c) (b^{2} + b c + c^{2})} \times x^{(c - a) (c^{2} + c a + a^{2})} = x^{(a^{3} - b^{3})} \times x^{(b^{3} - c^{3})} \times x^{(c^{3} - a^{3})} = x^{(a^{3} - b^{3} + b^{3} - c^{3} + c^{3} - a^{3})} = x^{0} = 1$
Left hand side is equal to right hand side.
Hence proved.

(ii) ${(\frac{x^{a}}{x^{b}})}^{c} \times {(\frac{x^{b}}{x^{c}})}^{a} \times {(\frac{x^{c}}{x^{a}})}^{b} = 1$
Consider the left hand side:
$= \frac{x^{a c}}{x^{b c}} \times \frac{x^{b a}}{x^{c a}} \times \frac{x^{c b}}{x^{a b}} = \frac{x^{a c}}{x^{b c}} \times \frac{x^{b a}}{x^{c a}} \times \frac{x^{c b}}{x^{a b}} = \frac{x^{a c} \times x^{b a} \times x^{c b}}{x^{b c} \times x^{c a} \times x^{a b}} = \frac{x^{a c + b a + c b}}{x^{b c + c a + a b}} = 1$
Left hand side is equal to right hand side.
Hence proved.

Page No 2.12:

Question 4:

Prove that:

(i) $\frac{1}{1 + x^{a - b}} + \frac{1}{1 + x^{b - a}} = 1$

(ii) $\frac{1}{1 + x^{b - a} + x^{c - a}} + \frac{1}{1 + x^{a - b} + x^{c - b}} + \frac{1}{1 + x^{b - c} + x^{a - c}} = 1$

Answer:

(i) Consider the left hand side:
$\frac{1}{1 + x^{a - b}} + \frac{1}{1 + x^{b - a}} = \frac{1}{1 + \frac{x^{a}}{x^{b}}} + \frac{1}{1 + \frac{x^{b}}{x^{a}}} = \frac{1}{\frac{x^{b} + x^{a}}{x^{b}}} + \frac{1}{\frac{x^{a} + x^{b}}{x^{a}}} = \frac{x^{b}}{x^{b} + x^{a}} + \frac{x^{a}}{x^{a} + x^{b}} = \frac{x^{b} + x^{a}}{x^{b} + x^{a}} = 1$
Therefore left hand side is equal to the right hand side. Hence proved.

(ii)
Consider the left hand side:
$\frac{1}{1 + x^{b - a} + x^{c - a}} + \frac{1}{1 + x^{a - b} + x^{c - b}} + \frac{1}{1 + x^{b - c} + x^{a - c}} = \frac{1}{1 + \frac{x^{b}}{x^{a}} + \frac{x^{c}}{x^{a}}} + \frac{1}{1 + \frac{x^{a}}{x^{b}} + \frac{x^{c}}{x^{b}}} + \frac{1}{1 + \frac{x^{b}}{x^{c}} + \frac{x^{a}}{x^{c}}} = \frac{1}{\frac{x^{a} + x^{b} + x^{c}}{x^{a}}} + \frac{1}{\frac{x^{b} + x^{a} + x^{c}}{x^{b}}} + \frac{1}{\frac{x^{c} + x^{b} + x^{c}}{x^{c}}} = \frac{x^{a}}{x^{a} + x^{b} + x^{c}} + \frac{x^{b}}{x^{b} + x^{a} + x^{c}} + \frac{x^{c}}{x^{c} + x^{b} + x^{c}} = \frac{x^{a} + x^{b} + x^{c}}{x^{a} + x^{b} + x^{c}} = 1$
Therefore left hand side is equal to the right hand side. Hence proved.

Page No 2.12:

Question 5:

Prove that:

(i) $\frac{a + b + c}{a^{- 1} b^{- 1} + b^{- 1} c^{- 1} + c^{- 1} a^{- 1}} = a b c$

(ii) ${(a^{- 1} + b^{- 1})}^{- 1} = \frac{a b}{a + b}$

Answer:

(i) Consider the left hand side:
$\frac{a + b + c}{a^{- 1} b^{- 1} + b^{- 1} c^{- 1} + c^{- 1} a^{- 1}} = \frac{a + b + c}{\frac{1}{a b} + \frac{1}{b c} + \frac{1}{c a}} = \frac{a + b + c}{\frac{c + a + b}{a b c}} = (a + b + c) \times (\frac{a b c}{a + b + c}) = a b c$
Therefore left hand side is equal to the right hand side. Hence proved.

(ii)
Consider the left hand side:
${(a^{- 1} + b^{- 1})}^{- 1} = \frac{1}{a^{- 1} + b^{- 1}} = \frac{1}{\frac{1}{a} + \frac{1}{b}} = \frac{1}{\frac{b + a}{a b}} = \frac{a b}{a + b}$
Therefore left hand side is equal to the right hand side. Hence proved.

Page No 2.12:

Question 6:

If abc = 1, show that $\frac{1}{1 + a + b^{- 1}} + \frac{1}{1 + b + c^{- 1}} + \frac{1}{1 + c + a^{- 1}} = 1$ .

Answer:

Consider the left hand side:
$\frac{1}{1 + a + b^{- 1}} + \frac{1}{1 + b + c^{- 1}} + \frac{1}{1 + c + a^{- 1}} = \frac{1}{1 + a + \frac{1}{b}} + \frac{1}{1 + b + \frac{1}{c}} + \frac{1}{1 + c + \frac{1}{a}} = \frac{1}{\frac{b + a b + 1}{b}} + \frac{1}{1 + b + a b} + \frac{1}{1 + \frac{1}{a b} + \frac{1}{a}} (a b c = 1) = \frac{b}{b + a b + 1} + \frac{1}{1 + b + a b} + \frac{a b}{a b + 1 + b} = \frac{b + 1 + a b}{b + a b + 1} = 1$
Hence proved.

Page No 2.12:

Question 7:

Simplify the following:

(i) $\frac{3^{n} \times 9^{n + 1}}{3^{n - 1} \times 9^{n - 1}}$

(ii) $\frac{5 \times 25^{n + 1} - 25 \times 5^{2 n}}{5 \times 5^{2 n + 3} - 25^{n + 1}}$

(iii) $\frac{5^{n + 3} - 6 \times 5^{n + 1}}{9 \times 5^{x} - 2^{2} \times 5^{n}}$

(iv) $\frac{6 {(8)}^{n + 1} + 16 {(2)}^{3 n - 2}}{10 {(2)}^{3 n + 1} - 7 {(8)}^{n}}$

Answer:

(i)
$\frac{3^{n} \times 9^{n + 1}}{3^{n - 1} \times 9^{n - 1}} = \frac{3^{n} \times {(3^{2})}^{(n + 1)}}{3^{n - 1} \times {(3^{2})}^{n - 1}} = \frac{3^{n} \times 3^{2 n + 2}}{3^{n - 1} \times 3^{2 n - 2}}$
$= \frac{3^{n + 2 n + 2}}{3^{n - 1 + 2 n - 2}} = \frac{3^{3 n + 2}}{3^{3 n - 3}} = 3^{3 n + 2 - 3 n + 3} = 3^{5} = 243$

(ii)
$\frac{5 \times 25^{n + 1} - 25 \times 5^{2 n}}{5 \times 5^{2 n + 3} - 25^{n + 1}} = \frac{5 \times {(5^{2})}^{n + 1} - (5^{2}) \times 5^{2 n}}{5 \times 5^{2 n + 3} - {(5^{2})}^{n + 1}} = \frac{5 \times (5^{2 n + 2}) - (5^{2}) \times 5^{2 n}}{5 \times 5^{2 n + 3} - (5^{2 n + 2})} = \frac{5^{1 + 2 n + 2} - 5^{2 + 2 n}}{5^{1 + 2 n + 3} - 5^{2 n + 2}}$
$= \frac{5^{2 n + 3} - 5^{2 + 2 n}}{5^{2 n + 4} - 5^{2 n + 2}} = \frac{5^{2 + 2 n} (5 - 1)}{5^{2 n + 2} (5^{2} - 1)} = \frac{4}{24} = \frac{1}{6}$

(iii)
$\frac{5^{n + 3} - 6 \times 5^{n + 1}}{9 \times 5^{n} - 2^{2} \times 5^{n}} = \frac{5^{n + 1} (5^{2} - 6)}{5^{n} (9 - 2^{2})} = \frac{5^{n} \times 5 \times (25 - 6)}{5^{n} (9 - 4)} = \frac{5 \times 19}{5} = 19$

(iv)
$\frac{6 {(8)}^{n + 1} + 16 {(2)}^{3 n - 2}}{10 {(2)}^{3 n + 1} - 7 {(8)}^{n}} = \frac{6 {(2^{3})}^{n + 1} + 16 {(2)}^{3 n - 2}}{10 {(2)}^{3 n + 1} - 7 {(2^{3})}^{n}} = \frac{6 (2^{3 n + 3}) + 16 {(2)}^{3 n - 2}}{10 {(2)}^{3 n + 1} - 7 (2^{3 n})}$
$= \frac{6 \times 2^{3 n} (2^{3}) + 16 {(2)}^{3 n} 2^{- 2}}{10 {(2)}^{3 n} (2^{1}) - 7 (2^{3 n})} = \frac{2^{3 n} (48 + 4)}{2^{3 n} (20 - 7)} = \frac{52}{13} = 4$

Page No 2.12:

Question 8:

Solve the following equations for x:

(i) $7^{2 x + 3} = 1$

(ii) $2^{x + 1} = 4^{x - 3}$

(iii) $2^{5 x + 3} = 8^{x + 3}$

(iv) $4^{2 x} = \frac{1}{32}$

(v) $4^{x - 1} \times {(0.5)}^{3 - 2 x} = {(\frac{1}{8})}^{x}$

(vi) $2^{3 x - 7} = 256$

Answer:

(i)
$7^{2 x + 3} = 1 \Rightarrow 7^{2 x + 3} = 7^{0} \Rightarrow 2 x + 3 = 0 \Rightarrow 2 x = - 3 \Rightarrow x = - \frac{3}{2}$

(ii)
$2^{x + 1} = 4^{x - 3} \Rightarrow 2^{x + 1} = {(2^{2})}^{x - 3} \Rightarrow 2^{x + 1} = (2^{2 x - 6}) \Rightarrow x + 1 = 2 x - 6 \Rightarrow x = 7$

(iii)
$2^{5 x + 3} = 8^{x + 3} \Rightarrow 2^{5 x + 3} = {(2^{3})}^{x + 3} \Rightarrow 2^{5 x + 3} = 2^{3 x + 9} \Rightarrow 5 x + 3 = 3 x + 9 \Rightarrow 2 x = 6 \Rightarrow x = 3$

(iv)
$4^{2 x} = \frac{1}{32} \Rightarrow {(2^{2})}^{2 x} = \frac{1}{2^{5}} \Rightarrow 2^{4 x} \times 2^{5} = 1 \Rightarrow 2^{4 x + 5} = 2^{0} \Rightarrow 4 x + 5 = 0 \Rightarrow x = - \frac{5}{4}$

(v)
$4^{x - 1} \times {(0.5)}^{3 - 2 x} = {(\frac{1}{8})}^{x} \Rightarrow {(2^{2})}^{x - 1} \times {(\frac{1}{2})}^{3 - 2 x} = {(\frac{1}{2^{3}})}^{x} \Rightarrow {(2^{2})}^{x - 1} \times {(2^{- 1})}^{3 - 2 x} = {(2^{- 3})}^{x} \Rightarrow 2^{2 x - 2} \times 2^{2 x - 3} = 2^{- 3 x} \Rightarrow 2^{2 x - 2 + 2 x - 3} = 2^{- 3 x} \Rightarrow 2^{4 x - 5} = 2^{- 3 x} \Rightarrow 4 x - 5 = - 3 x \Rightarrow 7 x = 5 \Rightarrow x = \frac{5}{7}$

(vi)
$2^{3 x - 7} = 256 \Rightarrow 2^{3 x - 7} = 2^{8} \Rightarrow 3 x - 7 = 8 \Rightarrow 3 x = 15 \Rightarrow x = 5$

Page No 2.12:

Question 9:

Solve the following equations for x:

(i) $2^{2 x} - 2^{x + 3} + 2^{4} = 0$

(ii) $3^{2 x + 4} + 1 = 2 . 3^{x + 2}$

Answer:

(i)
$2^{2 x} - 2^{x + 3} + 2^{4} = 0 \Rightarrow {(2^{x})}^{2} - (2^{x} \times 2^{3}) + {(2^{2})}^{2} = 0 \Rightarrow {(2^{x})}^{2} - 2 \times 2^{x} \times 2^{2} + {(2^{2})}^{2} = 0 \Rightarrow {(2^{x} - 2^{2})}^{2} = 0 \Rightarrow 2^{x} - 2^{2} = 0 \Rightarrow 2^{x} = 2^{2} \Rightarrow x = 2$

(ii)
$3^{2 x + 4} + 1 = 2 . 3^{x + 2} \Rightarrow {(3^{x + 2})}^{2} - 2 . 3^{x + 2} + 1 = 0 \Rightarrow {(3^{x + 2} - 1)}^{2} = 0 \Rightarrow 3^{x + 2} - 1 = 0 \Rightarrow 3^{x + 2} = 1 = 3^{0} \Rightarrow x + 2 = 0 \Rightarrow x = - 2$

Page No 2.13:

Question 10:

If $49392 = a^{4} b^{2} c^{3}$ , find the values of a, b and c, where a, b and c are different positive primes.

Answer:

First find out the prime factorisation of 49392.

$\begin{matrix} 2 & 49392 \\ 2 & 24696 \\ 2 & 12348 \\ 2 & 6174 \\ 3 & 3087 \\ 3 & 1029 \\ 7 & 343 \\ 7 & 49 \\ 7 & 7 \\ 1 \end{matrix}$

It can be observed that 49392 can be written as $2^{4} \times 3^{2} \times 7^{3}$ , where 2, 3 and 7 are positive primes.
$∴ 49392 = 2^{4} 3^{2} 7^{3} = a^{4} b^{2} c^{3} \Rightarrow a = 2, b = 3, c = 7$

Page No 2.13:

Question 11:

If $1176 = 2^{a} 3^{b} 7^{c}$ , find a, b and c.

Answer:

First find out the prime factorisation of 1176.
$\begin{matrix} 2 & 1176 \\ 2 & 588 \\ 2 & 294 \\ 3 & 147 \\ 7 & 49 \\ 7 & 7 \\ 1 \end{matrix}$

It can be observed that 1176 can be written as $2^{3} \times 3^{1} \times 7^{2}$ .
$1176 = 2^{3} 3^{1} 7^{2} = 2^{a} 3^{b} 7^{c}$
Hence, a = 3, b = 1 and c = 2.

Page No 2.13:

Question 12:

Given $4725 = 3^{a} 5^{b} 7^{c}$ , find
(i) the integral values of a, b and c
(ii) the value of $2^{- a} 3^{b} 7^{c}$

Answer:

(i) Given $4725 = 3^{a} 5^{b} 7^{c}$
First find out the prime factorisation of 4725.
$\begin{matrix} 3 & 4725 \\ 3 & 1575 \\ 3 & 525 \\ 5 & 175 \\ 5 & 35 \\ 7 & 7 \\ 1 \end{matrix}$
It can be observed that 4725 can be written as $3^{3} \times 5^{2} \times 7^{1}$ .
$∴ 4725 = 3^{a} 5^{b} 7^{c} = 3^{3} 5^{2} 7^{1}$
Hence, a = 3, b = 2 and c = 1.

(ii)
When a = 3, b = 2 and c = 1,
$2^{- a} 3^{b} 7^{c} = 2^{- 3} \times 3^{2} \times 7^{1} = \frac{1}{8} \times 9 \times 7 = \frac{63}{8}$
Hence, the value of $2^{- a} 3^{b} 7^{c}$ is $\frac{63}{8}$ .

Page No 2.13:

Question 13:

If $a = x y^{p - 1}, b = x y^{q - 1} and c = x y^{r - 1}$ , prove that $a^{q - r} b^{r - p} c^{p - q} = 1$ .

Answer:

It is given that $a = x y^{p - 1}, b = x y^{q - 1} and c = x y^{r - 1}$ .
$∴ a^{q - r} b^{r - p} c^{p - q} = {(x y^{p - 1})}^{q - r} {(x y^{q - 1})}^{r - p} {(x y^{r - 1})}^{p - q} = x^{(q - r)} y^{(p - 1) (q - r)} x^{(r - p)} y^{(r - p) (q - 1)} x^{(p - q)} y^{(p - q) (r - 1)} = x^{(q - r)} x^{(r - p)} x^{(p - q)} y^{(p - 1) (q - r)} y^{(r - p) (q - 1)} y^{(p - q) (r - 1)} = x^{(q - r) + (r - p) + (p - q)} y^{(p - 1) (q - r) + (r - p) (q - 1) + (p - q) (r - 1)} = x^{q - r + r - p + p - q} y^{p q - q - p r + r + r q - r - p q + p + p r - p - q r + q} = x^{0} y^{0} = 1$

Page No 2.24:

Question 1:

Assuming that x, y, z are positive real numbers, simplify each of the following:

(i) ${(\sqrt{x^{- 3}})}^{5}$

(ii) $\sqrt{x^{3} y^{- 2}}$

(iii) $(x^{- 2 / 3} y^{- 1 / 2})^{2}$

(iv) $(\sqrt{x})^{- 2 / 3} \sqrt{y^{4}} \div \sqrt{x y^{- 1 / 2}}$

(v) $\sqrt[5]{243 x^{10} y^{5} z^{10}}$

(vi) ${(\frac{x^{- 4}}{y^{- 10}})}^{5 / 4}$

(vii) ${(\frac{\sqrt{2}}{\sqrt{3}})}^{5} {(\frac{6}{7})}^{2}$

Answer:

We have to simplify the following, assuming thatare positive real numbers

(i) Given

As x is positive real number then we have

Hence the simplified value of is

(ii) Given

As x and y are positive real numbers then we can write

By using law of rational exponents we have

Hence the simplified value of is

(iii) Given

As x and y are positive real numbers then we have

By using law of rational exponents we have

${(x^{\frac{- 2}{3}} y^{^{\frac{- 1}{2}}})}^{2} = \frac{1}{x^{\frac{2}{3} + \frac{2}{3}}} \times \frac{1}{y^{\frac{1}{2} + \frac{1}{2}}} = \frac{1}{x^{\frac{4}{3}}} \times \frac{1}{y^{\frac{2}{2}}} = \frac{1}{x^{\frac{4}{3}}} \times \frac{1}{y} = \frac{1}{x^{\frac{4}{3}} y}$

Hence the simplified value of is .

$(iv) {(\sqrt{x})}^{- \frac{2}{3}} \sqrt{y^{4}} \div \sqrt{x y^{- \frac{1}{2}}} = {(x^{\frac{1}{2}})}^{- \frac{2}{3}} {(y^{4})}^{\frac{1}{2}} \div {(x \times y^{- \frac{1}{2}})}^{\frac{1}{2}} = \frac{x^{\frac{1}{2} \times - \frac{2}{3}} \times y^{4 \times \frac{1}{2}}}{x^{\frac{1}{2}} \times y^{- \frac{1}{2} \times \frac{1}{2}}} = \frac{x^{- \frac{1}{3}} \times y^{2}}{x^{\frac{1}{2}} \times y^{- \frac{1}{4}}} by using the law of rational exponents, a^{m} \div a^{n} = a^{m - n}, we have x^{- \frac{1}{3} - \frac{1}{2}} \times y^{2 + \frac{1}{4}} = x^{- \frac{5}{6}} \times y^{\frac{9}{4}} = \frac{y^{\frac{9}{4}}}{x^{\frac{5}{6}}} (v) . \sqrt[5]{243 x^{10} y^{5} z^{10}} = {(243 \times x^{10} \times y^{5} \times z^{10})}^{\frac{1}{5}} = {(243)}^{\frac{1}{5}} \times {(x^{10})}^{\frac{1}{5}} \times {(y^{5})}^{\frac{1}{5}} \times {(z^{10})}^{\frac{1}{5}} = {(3^{5})}^{\frac{1}{5}} \times x^{10 \times \frac{1}{5}} \times y^{5 \times \frac{1}{5}} \times z^{10 \times \frac{1}{5}} = 3 \times x^{2} \times y \times z^{2} = 3 x^{2} y z^{2} (vi) {(\frac{x^{- 4}}{y^{- 10}})}^{\frac{5}{4}} = \frac{{(x^{- 4})}^{\frac{5}{4}}}{{(y^{- 10})}^{\frac{5}{4}}} = \frac{x^{- 4 \times \frac{5}{4}}}{y^{- 10 \times \frac{5}{4}}} = \frac{x^{- 5}}{y^{- \frac{25}{2}}} = \frac{y^{\frac{25}{2}}}{x^{5}}$

(vii) ${(\frac{\sqrt{2}}{\sqrt{3}})}^{5} {(\frac{6}{7})}^{2}$

${(\frac{\sqrt{2}}{\sqrt{3}})}^{5} {(\frac{6}{7})}^{2} = {(\frac{\sqrt{2}}{\sqrt{3}})}^{2 + 2 + 1} {(\frac{6}{7})}^{2} = {(\frac{\sqrt{2}}{\sqrt{3}})}^{2} \times {(\frac{\sqrt{2}}{\sqrt{3}})}^{2} \times {(\frac{\sqrt{2}}{\sqrt{3}})}^{1} \times {(\frac{6}{7})}^{2} = (\frac{2}{3}) \times (\frac{2}{3}) \times {(\frac{\sqrt{2}}{\sqrt{3}})}^{1} \times {(\frac{6}{7})}^{2}$
$= (\frac{2}{3}) \times (\frac{2}{3}) \times {(\frac{\sqrt{2}}{\sqrt{3}})}^{1} \times {(\frac{6}{7})}^{2} = \frac{16 \sqrt{2}}{49 \sqrt{3}} = \sqrt{\frac{512}{7203}} = {(\frac{512}{7203})}^{\frac{1}{2}}$

Page No 2.24:

Question 2:

Simplify:

(i) $(16^{- 1 / 5})^{5 / 2}$

(ii) $\sqrt[5]{{(32)}^{- 3}}$

(iii) $\sqrt[3]{(343)^{- 2}}$

(iv) $(0.001)^{1 / 3}$

(v) $\frac{(25)^{3 / 2} \times (243)^{3 / 5}}{(16)^{5 / 4} \times (8)^{4 / 3}}$

(vi) ${(\frac{\sqrt{2}}{5})}^{8} \div {(\frac{\sqrt{2}}{5})}^{13}$

(vii) ${(\frac{5^{- 1} \times 7^{2}}{5^{2} \times 7^{- 4}})}^{7 / 2} \times {(\frac{5^{- 2} \times 7^{3}}{5^{3} \times 7^{- 5}})}^{- 5 / 2}$

Answer:

(1) Given

By using law of rational exponents we have

Hence the value of is
(ii) $\sqrt[5]{{(32)}^{- 3}}$

$= \sqrt[5]{{(\frac{1}{32})}^{3}} = {(\frac{1}{32})}^{\frac{3}{5}} = {(\frac{1}{{(2)}^{5}})}^{\frac{3}{5}} = {(\frac{1}{2})}^{5 \times \frac{3}{5}}$
$= {(\frac{1}{2})}^{3} = (\frac{1}{8})$

(iii) Given

Hence the value of is

(iv) Given

The value of is

(v) Given

Hence the value of is

(vi) Given. So,

By using the law of rational exponents

Hence the value of is

(vii) Given . So,

Hence the value of is

Page No 2.24:

Question 3:

Prove that:

(i) $\sqrt{3 \times 5^{- 3}} \div \sqrt[3]{3^{- 1}} \sqrt{5} \times \sqrt[6]{3 \times 5^{6}} = \frac{3}{5}$

(ii) $9^{3 / 2} - 3 \times 5^{0} - {(\frac{1}{81})}^{-}^{1 / 2} = 15$

(iii) ${(\frac{1}{4})}^{- 2} - 3 \times 8^{2 / 3} \times 4^{0} + {(\frac{9}{16})}^{- 1 / 2} = \frac{16}{3}$

(iv) $\frac{2^{1 / 2} \times 3^{1 / 3} \times 4^{1 / 4}}{10^{- 1 / 5} \times 5^{3 / 5}} \div \frac{3^{4 / 3} \times 5^{- 7 / 5}}{4^{- 3 / 5} \times 6} = 10$

(v) $\sqrt{\frac{1}{4} + (0.01)^{- 1 / 2}} - (27)^{2 / 3} = \frac{3}{2}$

(vi) $\frac{2^{n} + 2^{n - 1}}{2^{n + 1} - 2^{n}} = \frac{3}{2}$

(vii) ${(\frac{64}{125})}^{- 2 / 3} + \frac{1}{{(\frac{256}{625})}^{1 / 4}} + (\frac{\sqrt{25}}{\sqrt[3]{64}}) = \frac{65}{16}$

(viii) $\frac{3^{- 3} \times 6^{2} \times \sqrt{98}}{5^{2} \times \sqrt[3]{1 / 25} \times (15) -^{4 / 3} \times 3^{1 / 3}} = 28 \sqrt{2}$

(ix) $\frac{(0.6)^{0} - (0.1) - 1}{{(\frac{3}{8})}^{- 1} {(\frac{3}{2})}^{3} + {(- \frac{1}{3})}^{- 1}} = - \frac{3}{2}$

Answer:

(i) We have to prove that

By using rational exponent we get,

Hence,

(ii) We have to prove that. So,

Hence,

(iii) We have to prove that

Now,

Hence,

(iv) We have to prove that. So,

Let

Hence,

(v) We have to prove that

Let

Hence,

(vi) We have to prove that . So,

Let

Hence,

(vii) We have to prove that. So let

By taking least common factor we get

Hence,

(viii) We have to prove that. So,

Let

Hence,

(ix) We have to prove that. So,

Let

Hence,

Page No 2.25:

Question 4:

Show that:
(i) $\frac{1}{1 + x^{a - b}} + \frac{1}{1 + x^{b - a}} = 1$

(ii) ${[\{\frac{x^{a (a - b)}}{x^{a (a + b)}}\} \div \{\frac{x^{b (b - a)}}{x^{b (b + a)}}\}]}^{a + b} = 1$

(iii) ${(x^{\frac{1}{a - b}})}^{\frac{1}{a - c}} {(x^{\frac{1}{b - c}})}^{\frac{1}{b - a}} {(x^{\frac{1}{c - a}})}^{\frac{1}{c - b}} = 1$

(iv) ${(\frac{x^{a^{2} + b^{2}}}{x^{a b}})}^{a + b} {(\frac{x^{b^{2} + c^{2}}}{x^{b c}})}^{b + c} {(\frac{x^{c^{2} + a^{2}}}{x^{a c}})}^{a + c} = x^{2 (a^{3} + b^{3} + c^{3})}$

(v) ${(x^{a - b})}^{a + b} {(x^{b - c})}^{b + c} {(x^{c - a})}^{c + a} = 1$

(vi) ${\{{(x^{a - a^{- 1}})}^{\frac{1}{a - 1}}\}}^{\frac{a}{a + 1}} = x$

(vii) ${(\frac{a^{x + 1}}{a^{y + 1}})}^{x + y} {(\frac{a^{y + 2}}{a^{z + 2}})}^{y + z} {(\frac{a^{z + 3}}{a^{x + 3}})}^{z + x} = 1$

(viii) ${(\frac{3^{a}}{3^{b}})}^{a + b} {(\frac{3^{b}}{3^{c}})}^{b + c} {(\frac{3^{c}}{3^{a}})}^{c + a} = 1$

Answer:

(i)
$\frac{1}{1 + x^{a - b}} + \frac{1}{1 + x^{b - a}}$
$= \frac{1}{1 + \frac{x^{a}}{x^{b}}} + \frac{1}{1 + \frac{x^{b}}{x^{a}}} = \frac{x^{b}}{x^{b} + x^{a}} + \frac{x^{a}}{x^{a} + x^{b}} = \frac{x^{b} + x^{a}}{x^{a} + x^{b}} = 1$

(ii)
${[\{\frac{x^{a (a - b)}}{x^{a (a + b)}}\} \div \{\frac{x^{b (b - a)}}{x^{b (b + a)}}\}]}^{a + b} = {[\{\frac{x^{a (a - b)}}{x^{a (a + b)}}\} \times \{\frac{x^{b (b + a)}}{x^{b (b - a)}}\}]}^{a + b} = {[\{\frac{x^{a^{2} - a b}}{x^{a^{2} + a b}}\} \times \{\frac{x^{b^{2} + a b}}{x^{b^{2} - a b}}\}]}^{a + b}$
$= {[\{x^{a^{2} - a b - a^{2} - a b}\} \times \{x^{b^{2} + a b - b^{2} + a b}\}]}^{a + b} = {[x^{- 2 a b} \times x^{2 a b}]}^{a + b} = {[x^{- 2 a b + 2 a b}]}^{a + b} = {[x^{0}]}^{a + b} = 1$

(iii)
${(x^{\frac{1}{a - b}})}^{\frac{1}{a - c}} {(x^{\frac{1}{b - c}})}^{\frac{1}{b - a}} {(x^{\frac{1}{c - a}})}^{\frac{1}{c - b}} = {(x)}^{\frac{1}{a - b} \times \frac{1}{a - c}} {(x)}^{\frac{1}{b - c} \times}^{\frac{1}{b - a}} {(x)}^{\frac{1}{c - a} \times}^{\frac{1}{c - b}} = {(x)}^{\frac{1}{a - b} \times \frac{1}{a - c} + \frac{1}{b - c} \times \frac{1}{b - a} + \frac{1}{c - a} \times \frac{1}{c - b}} = {(x)}^{\frac{(b - c) - (a - c) + (a - b)}{(a - b) (a - c) (b - c)}} = x^{0} = 1$

(iv)
${(\frac{x^{a^{2} + b^{2}}}{x^{a b}})}^{a + b} {(\frac{x^{b^{2} + c^{2}}}{x^{b c}})}^{b + c} {(\frac{x^{c^{2} + a^{2}}}{x^{a c}})}^{a + c} = {(x^{a^{2} + b^{2} - a b})}^{a + b} {(x^{b^{2} + c^{2} - b c})}^{b + c} {(x^{c^{2} + a^{2} - a c})}^{a + c} = [x^{(a + b) (a^{2} + b^{2} - a b)}] [x^{(b + c) (b^{2} + c^{2} - b c)}] [x^{(a + c) (c^{2} + a^{2} - a c)}] = (x^{a^{3} + b^{3}}) (x^{b^{3} + c^{3}}) (x^{a^{3} + c^{3}}) = x^{2 (a^{3} + b^{3} + c^{3})}$

(v)
${(x^{a - b})}^{a + b} {(x^{b - c})}^{b + c} {(x^{c - a})}^{c + a} = [x^{(a - b) (a + b)}] [x^{(b - c) (b + c)}] [x^{(c - a) (c + a)}] = x^{(a^{2} - b^{2})} x^{(b^{2} - c^{2})} x^{(c^{2} - a^{2})} = x^{a^{2} - b^{2} + b^{2} - c^{2} + c^{2} - a^{2}} = x^{0} = 1$

(vi)
${\{{(x^{a - a^{- 1}})}^{\frac{1}{a - 1}}\}}^{\frac{a}{a + 1}} = \{{(x^{a - \frac{1}{a}})}^{\frac{1}{a - 1} \times \frac{a}{a + 1}}\} = {\{x^{\frac{a^{2} - 1}{a}}\}}^{\frac{a}{a^{2} - 1}} = x^{\frac{a^{2} - 1}{a} \times \frac{a}{a^{2} - 1}} = x^{1} = x$

(vii)
${(\frac{a^{x + 1}}{a^{y + 1}})}^{x + y} {(\frac{a^{y + 2}}{a^{z + 2}})}^{y + z} {(\frac{a^{z + 3}}{a^{x + 3}})}^{z + x} = {(a^{x + 1 - y - 1})}^{x + y} {(a^{y + 2 - z - 2})}^{y + z} {(a^{z + 3 - x - 3})}^{z + x} = {(a^{x - y})}^{x + y} {(a^{y - z})}^{y + z} {(a^{z - x})}^{z + x} = a^{x^{2} - y^{2} + y^{2} - z^{2} + z^{2} - x^{2}} = a^{0} = 1$

(viii)
${(\frac{3^{a}}{3^{b}})}^{a + b} {(\frac{3^{b}}{3^{c}})}^{b + c} {(\frac{3^{c}}{3^{a}})}^{c + a} = {(3^{a - b})}^{a + b} {(3^{b - c})}^{b + c} {(3^{c - a})}^{c + a} = 3^{a^{2} - b^{2} + b^{2} - c^{2} + c^{2} - a^{2}} = 3^{0} = 1$

Page No 2.25:

Question 5:

If $2^{x} = 3^{y} = 12^{z}, show that \frac{1}{z} = \frac{1}{y} + \frac{2}{x}$ .

Answer:

Let $2^{x} = 3^{y} = 12^{z} = k$
$\Rightarrow 2 = k^{\frac{1}{x}}, 3 = k^{\frac{1}{y}}, 12 = k^{\frac{1}{z}}$
Now,
$12 = k^{\frac{1}{z}} \Rightarrow 2^{2} \times 3 = k^{\frac{1}{z}} \Rightarrow {(k^{\frac{1}{x}})}^{2} \times k^{\frac{1}{y}} = k^{\frac{1}{z}} \Rightarrow k^{\frac{2}{x} + \frac{1}{y}} = k^{\frac{1}{z}} \Rightarrow \frac{2}{x} + \frac{1}{y} = \frac{1}{z}$

Page No 2.25:

Question 6:

If $2^{x} = 3^{y} = 6^{- z}, show that \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$ .

Answer:

Let $2^{x} = 3^{y} = 6^{- z} = k$
$\Rightarrow 2 = k^{\frac{1}{x}}, 3 = k^{\frac{1}{y}}, 6 = k^{\frac{1}{- z}}$
Now,
$6 = 2 \times 3 = k^{\frac{1}{- z}} \Rightarrow k^{\frac{1}{x}} \times k^{\frac{1}{y}} = k^{\frac{1}{- z}} \Rightarrow k^{\frac{1}{x} + \frac{1}{y}} = k^{\frac{1}{- z}} \Rightarrow \frac{1}{x} + \frac{1}{y} = \frac{1}{- z} \Rightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$

Page No 2.25:

Question 7:

If a^x= b^y= c^zand b² = ac, then show that $y = \frac{2 z x}{z + x}$ .

Answer:

Let $a^{x} = b^{y} = c^{z} = k$
So, $a = k^{\frac{1}{x}}, b = k^{\frac{1}{y}}, c = k^{\frac{1}{z}}$
Thus,
$b^{2} = a c \Rightarrow {(k^{\frac{1}{y}})}^{2} = (k^{\frac{1}{x}}) (k^{\frac{1}{z}}) \Rightarrow k^{\frac{2}{y}} = k^{\frac{1}{x} + \frac{1}{z}} \Rightarrow \frac{2}{y} = \frac{1}{x} + \frac{1}{z}$
$\Rightarrow \frac{2}{y} = \frac{z + x}{x z} \Rightarrow y = \frac{2 z x}{z + x}$

Page No 2.26:

Question 8:

If 3^x = 5^y = (75)^z, show that $z = \frac{x y}{2 x + y}$ .

Answer:

Let $3^{x} = 5^{y} = {(75)}^{z} = k$
$\Rightarrow 3 = k^{\frac{1}{x}}, 5 = k^{\frac{1}{y}}, 75 = k^{\frac{1}{z}} \Rightarrow 5^{2} \times 3 = k^{\frac{1}{z}} \Rightarrow {(k^{\frac{1}{y}})}^{2} \times k^{\frac{1}{x}} = k^{\frac{1}{z}} \Rightarrow k^{\frac{2}{y}} \times k^{\frac{1}{x}} = k^{\frac{1}{z}}$
$\Rightarrow k^{\frac{2}{y} + \frac{1}{x}} = k^{\frac{1}{z}} \Rightarrow \frac{2}{y} + \frac{1}{x} = \frac{1}{z} \Rightarrow \frac{2 x + y}{x y} = \frac{1}{z} \Rightarrow z = \frac{x y}{2 x + y}$

Page No 2.26:

Question 9:

If $27^{x} = \frac{9}{3^{x}},$ find x.

Answer:

We are given. We have to find the value of

Since

By using the law of exponents we get,

On equating the exponents we get,

Hence,

Page No 2.26:

Question 10:

Find the values of x in each of the following:

(i) $2^{5 x} \div 2 x = \sqrt[5]{2^{20}}$

(ii) $(2^{3})^{4} = (2^{2})^{x}$

(iii) ${(\frac{3}{5})}^{x} {(\frac{5}{3})}^{2 x} = \frac{125}{27}$

(iv) $5^{x - 2} \times 3^{2 x - 3} = 135$

(v) $2^{x - 7} \times 5^{x - 4} = 1250$

(vi) ${(\sqrt[3]{4})}^{2 x + \frac{1}{2}} = \frac{1}{32}$

(vii) $5^{2 x + 3} = 1$

(viii) ${(13)}^{\sqrt{x}} = 4^{4} - 3^{4} - 6$

(ix) ${(\sqrt{\frac{3}{5}})}^{x + 1} = \frac{125}{27}$

Answer:

From the following we have to find the value of x

(i) Given

By using rational exponents

On equating the exponents we get,

The value of x is

(ii) Given

On equating the exponents

Hence the value of x is

(iii) Given

Comparing exponents we have,

Hence the value of x is

(iv) Given

On equating the exponents of 5 and 3 we get,

And,

The value of x is

(v) Given

On equating the exponents we get

And,

Hence the value of x is

(vi) ${(\sqrt[3]{4})}^{2 x + \frac{1}{2}} = \frac{1}{32}$

${(2^{2})}^{\frac{1}{3} (\frac{4 x + 1}{2})} = {(\frac{1}{2})}^{5} \Rightarrow 2^{(\frac{4 x + 1}{3})} = 2^{- 5}$

On comparing we get,
$\frac{4 x + 1}{3} = - 5 \Rightarrow 4 x + 1 = - 15 \Rightarrow 4 x = - 16 \Rightarrow x = - 4$

(vii) $5^{2 x + 3} = 1$

$5^{2 x + 3} = 5^{0} \Rightarrow 2 x + 3 = 0 \Rightarrow x = \frac{- 3}{2}$

(viii) ${(13)}^{\sqrt{x}} = 4^{4} - 3^{4} - 6$

${(13)}^{\sqrt{x}} = {(2^{2})}^{4} - 3^{4} - 6 \Rightarrow {(13)}^{\sqrt{x}} = 2^{8} - 3^{4} - 6 \Rightarrow {(13)}^{\sqrt{x}} = 256 - 81 - 6 \Rightarrow {(13)}^{\sqrt{x}} = 169 \Rightarrow {(13)}^{\sqrt{x}} = {(13)}^{2}$

On comparing we get,
$\sqrt{x} = 2 on squaring both sides we get, x = 4$

(ix) ${(\sqrt{\frac{3}{5}})}^{x + 1} = \frac{125}{27}$
${(\sqrt{\frac{3}{5}})}^{x + 1} = {(\frac{5}{3})}^{3} \Rightarrow {(\frac{3}{5})}^{\frac{x + 1}{2}} = {(\frac{3}{5})}^{- 3}$

On comparing we get,

$\frac{x + 1}{2} = - 3 \Rightarrow x + 1 = - 6 \Rightarrow x = - 7$

Page No 2.26:

Question 11:

If $x = 2^{\frac{1}{3}} + 2^{\frac{2}{3}}$ , show that x³ – 6x = 6.

Answer:

$x = 2^{\frac{1}{3}} + 2^{\frac{2}{3}}$
Cubing on both sides, we get
$x^{3} = {(2^{\frac{1}{3}} + 2^{\frac{2}{3}})}^{3} \Rightarrow x^{3} = {(2^{\frac{1}{3}})}^{3} + {(2^{\frac{2}{3}})}^{3} + 3 \times 2^{\frac{1}{3}} \times 2^{\frac{2}{3}} (2^{\frac{1}{3}} + 2^{\frac{2}{3}}) \Rightarrow x^{3} = 2 + 2^{2} + 3 \times 2^{\frac{1 + 2}{3}} (x) \Rightarrow x^{3} = 2 + 4 + 3 \times 2 x \Rightarrow x^{3} - 6 x = 6$

Page No 2.26:

Question 12:

Determine ${(8 x)}^{x}, if 9^{x + 2} = 240 + 9^{x}$ .

Answer:

$9^{x + 2} = 240 + 9^{x}$
$\Rightarrow 9^{x} \times 9^{2} = 240 + 9^{x} \Rightarrow 9^{x} (9^{2} - 1) = 240 \Rightarrow 9^{x} (81 - 1) = 240 \Rightarrow 9^{x} \times 80 = 240 \Rightarrow 9^{x} = 3 \Rightarrow 3^{2 x} = 3^{1} \Rightarrow 2 x = 1 \Rightarrow x = \frac{1}{2}$
$∴ {(8 x)}^{x} = {(8 \times \frac{1}{2})}^{\frac{1}{2}} = {(4)}^{\frac{1}{2}} = 2$

Page No 2.26:

Question 13:

If $3^{x + 1} = 9^{x - 2}$ , find the value of 2^1+x.

Answer:

$3^{x + 1} = 9^{x - 2}$
$\Rightarrow 3^{x} \times 3 = \frac{9^{x}}{9^{2}} \Rightarrow 3^{x} \times 3 = \frac{{(3^{2})}^{x}}{{(3^{2})}^{2}} = \frac{3^{2 x}}{3^{4}} \Rightarrow 3^{4} \times 3 = \frac{3^{2 x}}{3^{x}} \Rightarrow 3^{5} = 3^{2 x - x}$
$\Rightarrow 3^{5} = 3^{x}$
Comparing both sides, we get
x = 5
So, $2^{1 + x} = 2^{1 + 5} = 2^{6} = 64$

Page No 2.26:

Question 14:

If $3^{4 x} = {(81)}^{- 1}$ and $10^{\frac{1}{y}} = 0.0001$ , find the value of $2^{- x + 4 y}$ .

Answer:

It is given that $3^{4 x} = {(81)}^{- 1}$ and $10^{\frac{1}{y}} = 0.0001$ .
Now,
$3^{4 x} = {(81)}^{- 1} \Rightarrow 3^{4 x} = {(3^{4})}^{- 1} \Rightarrow {(3^{x})}^{4} = {(3^{- 1})}^{4} \Rightarrow x = - 1$
And,
$10^{\frac{1}{y}} = 0.0001 \Rightarrow 10^{\frac{1}{y}} = \frac{1}{10000} \Rightarrow 10^{\frac{1}{y}} = {(\frac{1}{10})}^{4} \Rightarrow 10^{\frac{1}{y}} = {(10)}^{- 4} \Rightarrow \frac{1}{y} = - 4 \Rightarrow y = - \frac{1}{4}$
Therefore, the value of $2^{- x + 4 y}$ is $2^{1 + 4 (- \frac{1}{4})} = 2^{0} = 1$ .

Page No 2.26:

Question 15:

If $5^{3 x} = 125 and 10^{y} = 0.001$ , find x and y.

Answer:

It is given that $5^{3 x} = 125 and 10^{y} = 0.001$ .
Now,
$5^{3 x} = 125 \Rightarrow 5^{3 x} = 5^{3} \Rightarrow 3 x = 3 \Rightarrow x = 1$
And,
$10^{y} = 0.001 \Rightarrow 10^{y} = \frac{1}{1000} \Rightarrow 10^{y} = 10^{- 3} \Rightarrow y = - 3$
Hence, the values of x and y are 1 and −3, respectively.

Page No 2.26:

Question 16:

Solve the following equations:
(i) $3^{x + 1} = 27 \times 3^{4}$
(ii) $4^{2 x} = {(\sqrt[3]{16})}^{- \frac{6}{y}} = {(\sqrt{8})}^{2}$
(iii) $3^{x - 1} \times 5^{2 y - 3} = 225$
(iv) $8^{x + 1} = 16^{y + 2} and, {(\frac{1}{2})}^{3 + x} = {(\frac{1}{4})}^{3 y}$
(v) $4^{x - 1} \times {(0.5)}^{3 - 2 x} = {(\frac{1}{8})}^{x}$

(vi) $\sqrt{\frac{a}{b}} = {(\frac{b}{a})}^{1 - 2 x}$ , where a and b are distinct primes

Answer:

(i)
$3^{x + 1} = 27 \times 3^{4} \Rightarrow 3^{x + 1} = 3^{3} \times 3^{4} \Rightarrow 3^{x + 1} = 3^{3 + 4} \Rightarrow 3^{x + 1} = 3^{7} \Rightarrow x + 1 = 7 \Rightarrow x = 6$

(ii)
$4^{2 x} = {(\sqrt[3]{16})}^{- \frac{6}{y}} = {(\sqrt{8})}^{2} \Rightarrow 4^{2 x} = {(\sqrt{8})}^{2} and {(\sqrt[3]{16})}^{- \frac{6}{y}} = {(\sqrt{8})}^{2} \Rightarrow 4^{2 x} = (8^{\frac{1}{2} \times 2}) and (16^{\frac{1}{3} \times - \frac{6}{y}}) = (8^{\frac{1}{2} \times 2}) \Rightarrow 4^{2 x} = 8 and (16^{- \frac{2}{y}}) = 8 \Rightarrow 2^{4 x} = 2^{3} and (2^{- \frac{8}{y}}) = 2^{3} \Rightarrow 4 x = 3 and - \frac{8}{y} = 3 \Rightarrow x = \frac{3}{4} and y = - \frac{8}{3}$

(iii)
$3^{x - 1} \times 5^{2 y - 3} = 225 \Rightarrow 3^{x - 1} \times 5^{2 y - 3} = 3 \times 3 \times 5 \times 5 \Rightarrow 3^{x - 1} \times 5^{2 y - 3} = 3^{2} \times 5^{2} \Rightarrow x - 1 = 2 and 2 y - 3 = 2 \Rightarrow x = 3 and y = \frac{5}{2}$

(iv)
$8^{x + 1} = 16^{y + 2} and, {(\frac{1}{2})}^{3 + x} = {(\frac{1}{4})}^{3 y} \Rightarrow {(2^{3})}^{x + 1} = {(2^{4})}^{y + 2} and {(\frac{1}{2})}^{3 + x} = {(\frac{1}{2^{2}})}^{3 y} \Rightarrow {(2)}^{3 x + 3} = {(2)}^{4 y + 8} and {(\frac{1}{2})}^{3 + x} = {(\frac{1}{2})}^{6 y} \Rightarrow 3 x + 3 = 4 y + 8 and 3 + x = 6 y$
Now,
$3 + x = 6 y \Rightarrow x = 6 y - 3$
Putting x = 6y − 3 in $3 x - 4 y = 5$ , we get
$3 (6 y - 3) - 4 y = 5 \Rightarrow 18 y - 9 - 4 y = 5 \Rightarrow 14 y = 14 \Rightarrow y = 1$
Putting y = 1 in $x = 6 y - 3$ , we get
$x = 6 \times 1 - 3 = 3$

(v)
$4^{x - 1} \times {(0.5)}^{3 - 2 x} = {(\frac{1}{8})}^{x} \Rightarrow {(2^{2})}^{x - 1} \times {(\frac{1}{2})}^{3 - 2 x} = {(\frac{1}{2^{3}})}^{x} \Rightarrow {(2)}^{2 x - 2} \times {(2)}^{- (3 - 2 x)} = {(2)}^{- 3 x} \Rightarrow {(2)}^{2 x - 2 - 3 + 2 x} = {(2)}^{- 3 x} \Rightarrow 4 x - 5 = - 3 x \Rightarrow 7 x = 5 \Rightarrow x = \frac{5}{7}$

(vi)
$\sqrt{\frac{a}{b}} = {(\frac{b}{a})}^{1 - 2 x} \Rightarrow {(\frac{a}{b})}^{\frac{1}{2}} = {(\frac{a}{b})}^{- (1 - 2 x)} \Rightarrow \frac{1}{2} = - (1 - 2 x) \Rightarrow \frac{1}{2} = 2 x - 1 \Rightarrow \frac{3}{2} = 2 x \Rightarrow x = \frac{3}{4}$

Page No 2.26:

Question 17:

If a and b are distinct primes such that $\sqrt[3]{a^{6} b^{- 4}} = a^{x} b^{2 y}$ , find x and y.

Answer:

Given: $\sqrt[3]{a^{6} b^{- 4}} = a^{x} b^{2 y}$
$\sqrt[3]{a^{6} b^{- 4}} = a^{x} b^{2 y} \Rightarrow {(a^{6} b^{- 4})}^{\frac{1}{3}} = a^{x} b^{2 y} \Rightarrow a^{6 \times \frac{1}{3}} b^{- 4 \times \frac{1}{3}} = a^{x} b^{2 y} \Rightarrow a^{2} b^{- \frac{4}{3}} = a^{x} b^{2 y} \Rightarrow x = 2 and y = - \frac{2}{3}$

Page No 2.26:

Question 18:

If a and b are different positive primes such that

(i) ${(\frac{a^{- 1} b^{2}}{a^{2} b^{- 4}})}^{7} \div (\frac{a^{3} b^{- 5}}{a^{- 2} b^{3}}) = a^{x} b^{y}$ , find x and y.

(ii) ${(a + b)}^{- 1} (a^{- 1} + b^{- 1}) = a^{x} b^{y}$ , find x + y + 2.

Answer:

(i)
${(\frac{a^{- 1} b^{2}}{a^{2} b^{- 4}})}^{7} \div (\frac{a^{3} b^{- 5}}{a^{- 2} b^{3}}) = a^{x} b^{y} \Rightarrow (\frac{a^{- 7} b^{14}}{a^{14} b^{- 28}}) \div (\frac{a^{3} b^{- 5}}{a^{- 2} b^{3}}) = a^{x} b^{y} \Rightarrow (a^{- 7 - 14} b^{14 + 28}) \div (a^{3 + 2} b^{- 5 - 3}) = a^{x} b^{y} \Rightarrow (a^{- 21} b^{42}) \div (a^{5} b^{- 8}) = a^{x} b^{y} \Rightarrow a^{- 21 - 5} b^{42 + 8} = a^{x} b^{y} \Rightarrow a^{- 26} b^{50} = a^{x} b^{y} \Rightarrow x = - 26 and y = 50$

(ii)
${(a + b)}^{- 1} (a^{- 1} + b^{- 1}) = a^{x} b^{y} \Rightarrow \frac{1}{(a + b)} (\frac{1}{a} + \frac{1}{b}) = a^{x} b^{y} \Rightarrow \frac{1}{(a + b)} (\frac{a + b}{a b}) = a^{x} b^{y} \Rightarrow (\frac{1}{a b}) = a^{x} b^{y} \Rightarrow {(a b)}^{- 1} = a^{x} b^{y} \Rightarrow a^{- 1} b^{- 1} = a^{x} b^{y} \Rightarrow x = - 1 and y = - 1$
Therefore, the value of x + y + 2 is −1 −1 + 2 = 0.

Page No 2.26:

Question 19:

If $2^{x} \times 3^{y} \times 5^{z} = 2160$ , find x , y and z. Hence, compute the value of $3^{x} \times 2^{- y} \times 5^{- z}$ .

Answer:

Given: $2^{x} \times 3^{y} \times 5^{z} = 2160$
First, find out the prime factorisation of 2160.
$\begin{matrix} 2 & 2160 \\ 2 & 1080 \\ 2 & 540 \\ 2 & 270 \\ 3 & 135 \\ 3 & 45 \\ 3 & 15 \\ 5 & 5 \\ 1 \end{matrix}$
It can be observed that 2160 can be written as $2^{4} \times 3^{3} \times 5^{1}$ .
Also,
$2^{x} \times 3^{y} \times 5^{z} = 2^{4} \times 3^{3} \times 5^{1} \Rightarrow x = 4, y = 3, z = 1$
Therefore, the value of $3^{x} \times 2^{- y} \times 5^{- z}$ is $3^{4} \times 2^{- 3} \times 5^{- 1} = 81 \times \frac{1}{8} \times \frac{1}{5} = \frac{81}{40}$ .

Page No 2.26:

Question 20:

If $1176 = 2^{a} \times 3^{b} \times 7^{c}$ , find the values of a, b and c. Hence, compute the value of $2^{a} \times 3^{b} \times 7^{- c}$ as a fraction.

Answer:

First find the prime factorisation of 1176.
$\begin{matrix} 2 & 1176 \\ 2 & 588 \\ 2 & 294 \\ 3 & 147 \\ 7 & 49 \\ 7 & 7 \\ 1 \end{matrix}$
It can be observed that 1176 can be written as $2^{3} \times 3^{1} \times 7^{2}$ .
$1176 = 2^{a} 3^{b} 7^{c} = 2^{3} 3^{1} 7^{2}$
So, a = 3, b = 1 and c = 2.
Therefore, the value of $2^{a} \times 3^{b} \times 7^{- c}$ is $2^{3} \times 3^{1} \times 7^{- 2} = 8 \times 3 \times \frac{1}{49} = \frac{24}{49}$

Page No 2.27:

Question 21:

Simplify:

(i) ${(\frac{x^{a + b}}{x^{c}})}^{a - b} {(\frac{x^{b + c}}{x^{a}})}^{b - c} {(\frac{x^{c + a}}{x^{b}})}^{c - a}$

(ii) $l m \sqrt{\frac{x^{l}}{x^{m}}} \times m n \sqrt{\frac{x^{m}}{x^{n}}} \times n l \sqrt{\frac{x^{n}}{x^{l}}}$

Answer:

(i)
${(\frac{x^{a + b}}{x^{c}})}^{a - b} {(\frac{x^{b + c}}{x^{a}})}^{b - c} {(\frac{x^{c + a}}{x^{b}})}^{c - a} = (\frac{x^{(a + b) (a - b)}}{x^{c (a - b)}}) (\frac{x^{(b + c) (b - c)}}{x^{a (b - c)}}) (\frac{x^{(c + a) (c - a)}}{x^{b (c - a)}}) = (\frac{x^{(a + b) (a - b)}}{x^{c a - b c}}) (\frac{x^{(b + c) (b - c)}}{x^{a b - a c}}) (\frac{x^{(c + a) (c - a)}}{x^{b c - a b}}) = (\frac{x^{(a^{2} - b^{2})} x^{(b^{2} - c^{2})} x^{(c^{2} - a^{2})}}{x^{c a - b c + a b - a c + b c - a b}}) = \frac{x^{a^{2} - b^{2} + b^{2} - c^{2} + c^{2} - a^{2}}}{x^{c a - b c + a b - a c + b c - a b}} = \frac{x^{0}}{x^{0}} = 1$

(ii)
$\sqrt[l m]{\frac{x^{l}}{x^{m}}} \times \sqrt[m n]{\frac{x^{m}}{x^{n}}} \times \sqrt[n l]{\frac{x^{n}}{x^{l}}} = {(\frac{x^{l}}{x^{m}})}^{\frac{1}{m l}} \times {(\frac{x^{m}}{x^{n}})}^{\frac{1}{m n}} \times {(\frac{x^{n}}{x^{l}})}^{\frac{1}{n l}} = {(x^{l - m})}^{\frac{1}{m l}} \times {(x^{m - n})}^{\frac{1}{m n}} \times {(x^{n - l})}^{\frac{1}{n l}} = x^{\frac{l - m}{m l}} \times x^{\frac{m - n}{m n}} \times x^{\frac{n - l}{n l}} = x^{\frac{l - m}{m l} + \frac{m - n}{m n} + \frac{n - l}{n l}} = x^{\frac{l n - m n + l m - n l + n m - l m}{n m l}} = x^{0} = 1$

Page No 2.27:

Question 22:

Show that: $\frac{{(a + \frac{1}{b})}^{m} \times {(a - \frac{1}{b})}^{n}}{{(b + \frac{1}{a})}^{m} \times {(b - \frac{1}{a})}^{n}} = {(\frac{a}{b})}^{m + n}$

Answer:

$\frac{{(a + \frac{1}{b})}^{m} \times {(a - \frac{1}{b})}^{n}}{{(b + \frac{1}{a})}^{m} \times {(b - \frac{1}{a})}^{n}} = \frac{{(\frac{a b + 1}{b})}^{m} \times {(\frac{a b - 1}{b})}^{n}}{{(\frac{a b + 1}{a})}^{m} \times {(\frac{a b - 1}{a})}^{n}} = {(\frac{\frac{a b + 1}{b}}{\frac{a b + 1}{a}})}^{m} \times {(\frac{\frac{a b - 1}{b}}{\frac{a b - 1}{a}})}^{n} = {(\frac{a b + 1}{b} \times \frac{a}{a b + 1})}^{m} \times {(\frac{a b - 1}{b} \times \frac{a}{a b - 1})}^{n} = {(\frac{a}{b})}^{m} \times {(\frac{a}{b})}^{n} = {(\frac{a}{b})}^{m} \times {(\frac{a}{b})}^{n} = {(\frac{a}{b})}^{m + n}$

Page No 2.27:

Question 23:

(i) If $a = x^{m + n} y^{l}, b = x^{n + l} y^{m} and c = x^{l + m} y^{n}$ , prove that $a^{m - n} b^{n - l} c^{l - m} = 1$ .

(ii) If $x = a^{m + n}, y = a^{n + l} and z = a^{l + m}$ , prove that $x^{m} y^{n} z^{l} = x^{n} y^{l} z^{m}$ .

Answer:

(i) Given: $a = x^{m + n} y^{l}, b = x^{n + l} y^{m} and c = x^{l + m} y^{n}$
Putting the values of a, b and c in $a^{m - n} b^{n - l} c^{l - m}$ , we get
$a^{m - n} b^{n - l} c^{l - m} = {(x^{m + n} y^{l})}^{m - n} {(x^{n + l} y^{m})}^{n - l} {(x^{l + m} y^{n})}^{l - m} = [x^{(m + n) (m - n)} y^{l (m - n)}] [x^{(n + l) (n - l)} y^{m (n - l)}] [x^{(l + m) (l - m)} y^{n (l - m)}] = x^{(m^{2} - n^{2})} x^{(n^{2} - l^{2})} x^{(l^{2} - m^{2})} y^{l m - l n} y^{m n - m l} y^{n l - n m} = x^{m^{2} - n^{2} + n^{2} - l^{2} + l^{2} - m^{2}} y^{l m - l n + m n - m l + n l - n m} = x^{0} y^{0} = 1$

(ii) Given: $x = a^{m + n}, y = a^{n + l} and z = a^{l + m}$
Putting the values of x, y and z in $x^{m} y^{n} z^{l}$ , we get
$x^{m} y^{n} z^{l} = {(a^{m + n})}^{m} {(a^{n + l})}^{n} {(a^{l + m})}^{l} = (a^{m^{2} + n m}) (a^{n^{2} + l n}) (a^{l^{2} + l m}) = a^{m^{2} + n^{2} + l^{2} + n m + l n + l m}$
Putting the values of x, y and z in $x^{n} y^{l} z^{m}$ , we get
$x^{n} y^{l} z^{m} = {(a^{m + n})}^{n} {(a^{n + l})}^{l} {(a^{l + m})}^{m} = (a^{m n + n^{2}}) (a^{n l + l^{2}}) (a^{l m + m^{2}}) = a^{m n + n^{2} + n l + l^{2} + l m + m^{2}}$
So, $x^{m} y^{n} z^{l}$ = $x^{n} y^{l} z^{m}$

Page No 2.28:

Question 1:

The value of ${\{2 - 3 (2 - 3)^{3}\}}^{3}$ is

(a) 5

(b) 125

(c) 1/5

(d) -125

Answer:

We have to find the value of. So,

The value of is 125

Hence the correct choice is

Page No 2.28:

Question 2:

The value of x − y^x-y when x = 2 and y = −2 is

(a) 18
(b) −18
(c) 14
(d) −14

Answer:

Given

Here

By substituting in we get

The value of is – 14

Hence the correct choice is .

Page No 2.28:

Question 3:

The product of the square root of x with the cube root of x is
(a) cube root of the square root of x
(b) sixth root of the fifth power of x
(c) fifth root of the sixth power of x
(d) sixth root of x

Answer:

We have to find the product (say L) of the square root of x with the cube root of x is. So,

$= x^{\frac{3 + 2}{6}} = x^{\frac{5}{6}}$

The product of the square root of x with the cube root of x is

Hence the correct alternative is

Page No 2.29:

Question 4:

The seventh root of x divided by the eighth root of x is
(a) x

(b) $\sqrt{x}$

(c) $\sqrt[56]{x}$

(d) $\frac{1}{\sqrt[56]{x}}$

Answer:

We have to find he seventh root of x divided by the eighth root of x, so let it be L. So,

The seventh root of x divided by the eighth root of x is

Hence the correct choice is .

Page No 2.29:

Question 5:

The square root of 64 divided by the cube root of 64 is

(a) 64
(b) 2
(c) $\frac{1}{2}$
(d) 64^2/3

Answer:

We have to find the value of .

So,

The value of is

Hence the correct choice is .

Page No 2.29:

Question 6:

The value of ${\{{(23 + 2^{2})}^{2 / 3} + (140 - 29)^{1 / 2}\}}^{2}$ , is

(a) 400

(b) 324

(c) 289

(d) 196

Answer:

Disclaimer: In question in place of 29 it should be 19.

We have to find the value of ${\{{(23 + 2^{2})}^{2 / 3} + (140 - 19)^{1 / 2}\}}^{2}$

$\Rightarrow {\{{(23 + 2^{2})}^{2 / 3} + (140 - 19)^{1 / 2}\}}^{2} = 20^{2} = 400$

Hence the correct answer is option (a).

Page No 2.29:

Question 7:

When simplified $(x^{- 1} + y^{- 1})^{- 1}$ is equal to

(a) xy

(b) x+y

(c) $\frac{x y}{x + y}$

(d) $\frac{x + y}{x y}$

Answer:

We have to simplify

So,

The value of is

Hence the correct choice is .

Page No 2.29:

Question 8:

If $8^{x + 1}$ = 64 , what is the value of $3^{2 x + 1}$ ?

(a) 1
(b) 3
(c) 9
(d) 27

Answer:

We have to find the value of provided

So,

Equating the exponents we get

By substitute in we get

The real value of is

Hence the correct choice is .

Page No 2.29:

Question 9:

If (2³)² = 4^x, then 3^x =

(a) 3
(b) 6
(c) 9
(d) 27

Answer:

We have to find the value ofprovided

So,

By equating the exponents we get

By substituting in we get

The value of is

Hence the correct choice is

Page No 2.29:

Question 10:

If x^{-2 =}64, then x¹^/3+x⁰ =

(a) 2
(b) 3
(c) 3/2
(d) 2/3

Answer:

We have to find the value ofif

Consider,

Multiply on both sides of powers we get

By taking reciprocal on both sides we get,

Substituting in we get

By taking least common multiply we get

Hence the correct choice is .

Page No 2.29:

Question 11:

When simplified ${(- \frac{1}{27})}^{- 2 / 3}$ is

(a) 9

(b) −9

(c) $\frac{1}{9}$

(d) $- \frac{1}{9}$

Answer:

We have to find the value of

So,

Hence the correct choice is .

Page No 2.29:

Question 12:

Which one of the following is not equal to ${(\sqrt[3]{8})}^{- 1 / 2} ?$

(a) ${\sqrt[3]{2}}^{- 1 / 2}$

(b) $8^{- 1 / 6}$

(c) $\frac{1}{(\sqrt[3]{8})^{1 / 2}}$

(d) $\frac{1}{\sqrt{2}}$

Answer:

We have to find the value of

So,

Also,

Hence the correct alternative is .

Page No 2.29:

Question 13:

Which one of the following is not equal to ${(\frac{100}{9})}^{- 3 / 2}$ ?

(a) ${(\frac{9}{100})}^{3 / 2}$

(b) ${(\frac{1}{\frac{100}{9}})}^{3 / 2}$

(c) $\frac{3}{10} \times \frac{3}{10} \times \frac{3}{10}$

(d) $\sqrt{\frac{100}{9}} \times \sqrt{\frac{100}{9}} \times \sqrt{\frac{100}{9}}$

Answer:

We have to find the value of

So,

Since, is equal to ,,.

Hence the correct choice is

Page No 2.29:

Question 14:

If a, b, c are positive real numbers, then $\sqrt{a^{- 1} b} \times \sqrt{b^{- 1} c} \times \sqrt{c^{- 1} a}$ is equal to

(a) 1

(b) abc

(c) $\sqrt{a b c}$

(d) $\frac{1}{a b c}$

Answer:

We have to find the value of when a, b, c are positive real numbers.

So,

Taking square root as common we get

$\sqrt{a^{- 1} b} \times \sqrt{b^{- 1} c} \times \sqrt{c^{- 1} a} = \sqrt{\frac{b}{a} \times \frac{c}{b} \times \frac{a}{c}} \sqrt{a^{- 1} b} \times \sqrt{b^{- 1} c} \times \sqrt{c^{- 1} a} = 1$

Hence the correct alternative is .

Page No 2.30:

Question 15:

, then x =

(a) 2
(b) 3
(c) 4
(d) 1

Answer:

We have to find value of provided

So,

Equating exponents of power we get

Hence the correct alternative is

Page No 2.30:

Question 16:

The value of ${\{8^{- 4 / 3} \div 2^{- 2}\}}^{1 / 2}$ is

(a) $\frac{1}{2}$

(b) 2

(c) $\frac{1}{4}$

(d) 4

Answer:

Find the value of

Hence the correct choice is .

Page No 2.30:

Question 17:

If a, b, c are positive real numbers, then $\sqrt[5]{3125 a^{10} b^{5} c^{10}}$ is equal to

(a) 5a²bc²

(b) 25ab²c

(c) 5a³bc³

(d) 125a²bc²

Answer:

Find value of.

$\sqrt[5]{3125 a^{10} b^{5} c^{10}} = 5 a^{2} b c^{2}$

Hence the correct choice is .

Page No 2.30:

Question 18:

If a, m, n are positive ingegers, then ${\{\sqrt[m]{\sqrt[n]{a}}\}}^{m n}$ is equal to

(a) a^mn

(b) a

(c) a^{m/n

(d) 1}

Answer:

Find the value of .

So,

Hence the correct choice is

Page No 2.30:

Question 19:

If x = 2 and y = 4, then ${(\frac{x}{y})}^{x - y} + {(\frac{y}{x})}^{y - x} =$

(a) 4

(b) 8

(c) 12

(d) 2

Answer:

We have to find the value of if,

Substitute,into get,

Hence the correct choice is .

Page No 2.30:

Question 20:

The value of m for which ${[{\{{(\frac{1}{7^{2}})}^{- 2}\}}^{- 1 / 3}]}^{1 / 4} = 7^{m},$ is

(a) $- \frac{1}{3}$

(b) $\frac{1}{4}$

(c) −3

(d) 2

Answer:

We have to find the value of for

By using rational exponents

$7^{\frac{- 1}{3}} = 7^{m}$

Equating power of exponents we get

Hence the correct choice is .

Page No 2.30:

Question 21:

The value of (0.00243)^3/5 + (0.0256)^3/4 is
(a) 0.083
(b) 0.073
(c) 0.081
(d) 0.091

Answer:

$\begin{array}{rcl} {(0.00243)}^{\frac{3}{5}} + {(0.0256)}^{\frac{3}{4}} & = & {(\frac{243}{100000})}^{\frac{3}{5}} + {(\frac{256}{10000})}^{\frac{3}{4}} \\ = & {(243 \times 10^{- 5})}^{\frac{3}{5}} + {(256 \times 10^{- 4})}^{\frac{3}{4}} \\ = & {(3^{5} \times 10^{- 5})}^{\frac{3}{5}} + {(4^{4} \times 10^{- 4})}^{\frac{3}{4}} \\ = & 3^{5 \times \frac{3}{5}} \times 10^{- 5 \times \frac{3}{5}} + 4^{4 \times \frac{3}{4}} \times 10^{- 4 \times \frac{3}{4}} \\ = & 3^{3} \times 10^{- 3} + 4^{3} \times 10^{- 3} \\ = & 27 \times 10^{- 3} + 64 \times 10^{- 3} \\ = & 0.027 + 0.064 \\ = & 0.091 \end{array}$

Hence, the correct answer is option (d).

Page No 2.30:

Question 22:

(256)^0.16 × (256)^0.09

(a) 4
(b) 16
(c) 64
(d) 256.25

Answer:

We have to find the value of. So,

By using law of rational exponents

we get

The value of is 4

Hence the correct choice is .

Page No 2.30:

Question 23:

If 10^2y= 25, then 10^-y equals

(a) $- \frac{1}{5}$

(b) $\frac{1}{50}$

(c) $\frac{1}{625}$

(d) $\frac{1}{5}$

Answer:

We have to find the value of

Given that, therefore,

Hence the correct option is .

Page No 2.30:

Question 24:

If 9^x⁺² = 240 + 9^x, then x =

(a) 0.5
(b) 0.2
(c) 0.4
(d) 0.1

Answer:

We have to find the value of

Given

By equating the exponents we get

Hence the correct alternative is .

Page No 2.30:

Question 25:

If x is a positive real number and x² = 2, then x³ =

(a) $\sqrt{2}$

(b) 2 $\sqrt{2}$

(c) 3 $\sqrt{2}$

(d) 4

Answer:

We have to find provided. So,

By raising both sides to the power

By substituting in we get

The value of is

Hence the correct choice is .

Page No 2.31:

Question 26:

If $\frac{x}{x^{1.5}} = 8 x^{- 1}$ and x > 0, then x =

(a) $\frac{\sqrt{2}}{4}$

(b) $2 \sqrt{2}$

(c) 4

(d) 64

Answer:

For, we have to find the value of x.

So,

By raising both sides to the power we get

The value of is

Hence the correct alternative is

Page No 2.31:

Question 27:

If $g = t^{2 / 3} + 4 t^{- 1 / 2}$ , What is the value of g when t = 64?

(a) $\frac{31}{2}$

(b) $\frac{33}{2}$

(c) $16$

(d) $\frac{257}{16}$

Answer:

Given.We have to find the value of

So,

The value of is

Hence the correct choice is

Page No 2.31:

Question 28:

If $4^{x} - 4^{x - 1} = 24,$ then (2x)^x equals

(a) $5 \sqrt{5}$

(b) $\sqrt{5}$

(c) $25 \sqrt{5}$

(d) 125

Answer:

We have to find the value of if

So,

Taking as common factor we get

By equating powers of exponents we get

By substituting in we get

Hence the correct choice is

Page No 2.31:

Question 29:

When simplified $(256)^{- (4^{- 3 / 2})}$ is

(a) 8

(b) $\frac{1}{8}$

(c) 2

(d) $\frac{1}{2}$

Answer:

Simplify

${(256)}^{-^{(4^{- \frac{3}{2}})}} = {(256)}^{- {(2)}^{(- 3)}}$

Hence the correct choice is .

Page No 2.31:

Question 30:

If $\frac{3^{2 x - 8}}{225} = \frac{5^{3}}{5^{x}},$ then x =

(a) 2
(b) 3
(c) 5
(d) 4

Answer:

We have to find the value of provided

So,

By cross multiplication we get

By equating exponents we get

And

Hence the correct choice is

Page No 2.31:

Question 31:

The value of 64^-1/3 (64^1/3-64^2/3), is

(a) 1

(b) $\frac{1}{3}$

(c) −3

(d) −2

Answer:

Find the value of

So,

Hence the correct statement is.

Page No 2.31:

Question 32:

If $\sqrt{5^{n}} = 125$ , then =

(a) 25

(b) $\frac{1}{125}$

(c) 625

(d) $\frac{1}{5}$

Answer:

We have to find provided

So,

Substitute in to get

Hence the value of is

The correct choice is

Page No 2.31:

Question 33:

If (16)^2x+3 =(64)^x+3, then 4^2x-2 =

(a) 64

(b) 256

(c) 32

(d) 512

Answer:

We have to find the value ofprovided

So,

Equating the power of exponents we get

The value of is

Hence the correct alternative is

Page No 2.31:

Question 34:

If $2^{- m} \times \frac{1}{2^{m}} = \frac{1}{4},$ then $\frac{1}{14} \{(4^{m})^{1 / 2} + {(\frac{1}{5^{m}})}^{- 1}\}$ is equal to

(a) $\frac{1}{2}$

(b) 2

(c) 4

(d) $- \frac{1}{4}$

Answer:

We have to find the value ofprovided

Consider,

Equating the power of exponents we get

By substituting we get

Hence the correct choice is .

Page No 2.31:

Question 35:

If $\frac{2^{m + n}}{2^{n - m}} = 16$ , $\frac{3^{p}}{3^{n}} = 81$ and $a = 2^{1 / 10}$ , then $\frac{a^{2 m + n - p}}{(a^{m - 2 n + 2 p})^{- 1}} =$

(a) 2

(b) $\frac{1}{4}$

(c) 9

(d) $\frac{1}{8}$

Answer:

Given : , $\frac{3^{p}}{3^{n}} = 81$ and
To find :

Find :
By using rational components We get

By equating rational exponents we get

Now, = $(a^{2 m + n - p}) . (a^{m - 2 n + 2 p})$ we get
$= a^{2 m + n - p + m - 2 n + 2 p} = a^{3 m - n + p} Now putting value of a = 2^{\frac{1}{10}} we get, = 2^{\frac{3 m - n + p}{10}} = 2^{\frac{6 - n + p}{10}}$

Also, $\frac{3^{p}}{3^{n}} = 81$
$3^{p - n} = 3^{4}$
On comparing LHS and RHS we get, p - n = 4.
Now,
= a^{3m - n + p

$= 2^{\frac{6 + (p - n)}{10}} = 2^{\frac{6 + 4}{10}} = 2^{\frac{10}{10}} = 2^{1} = 2$}

So, option (a) is the correct answer.

Page No 2.32:

Question 36:

If $\frac{3^{5 x} \times 81^{2} \times 6561}{3^{2 x}} = 3^{7}$ , then x =

(a) 3

(b) −3

(c) $\frac{1}{3}$

(d) $- \frac{1}{3}$

Answer:

We have to find the value of x provided

So,

By using law of rational exponents we get

By equating exponents we get

Hence the correct choice is .

Page No 2.32:

Question 37:

If 0 < y < x, which statement must be true?
(a) $\sqrt{x} - \sqrt{y} = \sqrt{x - y}$
(b) $\sqrt{x} + \sqrt{x} = \sqrt{2 x}$
(c) $x \sqrt{y} = y \sqrt{x}$
(d) $\sqrt{x y} = \sqrt{x} \sqrt{y}$

Answer:

Given
Option (a) :
Left hand side:

Right Hand side:

Left hand side is not equal to right hand side

The statement is wrong.

Option (b) :

Left hand side:

Right Hand side:

Left hand side is not equal to right hand side

The statement is wrong.

Option (c) :

Left hand side:

Right Hand side:

Left hand side is not equal to right hand side

The statement is wrong.

Option (d) :

Left hand side:

Right Hand side:

Left hand side is equal to right hand side

The statement is true.

Hence the correct choice is .

Page No 2.32:

Question 38:

If 10^x = 64, what is the value of $10^{\frac{x}{2} + 1} ?$

(a) 18
(b) 42
(c) 80
(d) 81

Answer:

We have to find the value of provided

So,

By substituting we get

Hence the correct choice is .

Page No 2.32:

Question 39:

$\frac{5^{n + 2} - 6 \times 5^{n + 1}}{13 \times 5^{n} - 2 \times 5^{n + 1}}$ is equal to

(a) $\frac{5}{3}$

(b) $- \frac{5}{3}$

(c) $\frac{3}{5}$

(d) $- \frac{3}{5}$

Answer:

We have to simplify

Taking as a common factor we get

Hence the correct alternative is

Page No 2.32:

Question 40:

If $\sqrt{2^{n}} = 1024,$ then ${3^{2}}^{(\frac{n}{4} - 4)} =$

(a) 3

(b) 9

(c) 27

(d) 81

Answer:

We have to find

Given

Equating powers of rational exponents we get

Substituting in we get

Hence the correct choice is .

Page No 2.32:

Question 1:

(21² – 15²)^2/3 is equal to ________.

Answer:

${(21^{2} - 15^{2})}^{\frac{2}{3}} = {(441 - 225)}^{\frac{2}{3}} = {(216)}^{\frac{2}{3}} = {(6^{3})}^{\frac{2}{3}} = {(6)}^{3 \times \frac{2}{3}} = {(6)}^{2} = 36$

Hence, (21² – 15²)^2/3 is equal to 36.

Page No 2.32:

Question 2:

81^1/4 × 9^3/2 × 27^–4/3 is equal to _________.

Answer:

$81^{\frac{1}{4}} \times 9^{\frac{3}{2}} \times 27^{- \frac{4}{3}} = {(3^{4})}^{\frac{1}{4}} \times {(3^{2})}^{\frac{3}{2}} \times \frac{1}{27^{\frac{4}{3}}} = {(3)}^{4 \times \frac{1}{4}} \times {(3)}^{2 \times \frac{3}{2}} \times \frac{1}{{(3^{3})}^{\frac{4}{3}}} = {(3)}^{1} \times {(3)}^{3} \times \frac{1}{{(3)}^{3 \times \frac{4}{3}}} = {(3)}^{1 + 3} \times \frac{1}{3^{4}} = {(3)}^{4} \times \frac{1}{3^{4}} = 1$

Hence, 81^1/4 × 9^3/2 × 27^–4/3 is equal to 1.

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Question 3:

${\{\frac{{(169)}^{- 3}}{{(196)}^{- 8}}\}}^{\frac{1}{48}} =$ __________.

Answer:

${\{\frac{{(169)}^{- 3}}{{(196)}^{- 8}}\}}^{\frac{1}{48}} = {\{\frac{{(13^{2})}^{- 3}}{{(14^{2})}^{- 8}}\}}^{\frac{1}{48}} = {\{\frac{{(13)}^{2 \times (- 3)}}{{(14)}^{2 \times (- 8)}}\}}^{\frac{1}{48}} = {\{\frac{{(13)}^{- 6}}{{(14)}^{- 16}}\}}^{\frac{1}{48}} = \frac{{(13)}^{- 6 \times \frac{1}{48}}}{{(14)}^{- 16 \times \frac{1}{48}}} = \frac{{(13)}^{- \frac{1}{8}}}{{(14)}^{- \frac{1}{3}}} = \frac{\frac{1}{13^{\frac{1}{8}}}}{\frac{1}{14^{\frac{1}{3}}}} = \frac{{(14)}^{\frac{1}{3}}}{{(13)}^{\frac{1}{8}}}$

Hence, ${\{\frac{{(169)}^{- 3}}{{(196)}^{- 8}}\}}^{\frac{1}{48}} =$ $\frac{{(14)}^{\frac{1}{3}}}{{(13)}^{\frac{1}{8}}}$ .

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Question 4:

If x = 8^2/3 × 32^–2/5, then x^–5= ________.

Answer:

$Let x = 8^{\frac{2}{3}} \times 32^{- \frac{2}{5}} \Rightarrow x = {(2^{3})}^{\frac{2}{3}} \times {(2^{5})}^{- \frac{2}{5}} \Rightarrow x = {(2)}^{3 \times \frac{2}{3}} \times {(2)}^{5 \times (- \frac{2}{5})} \Rightarrow x = {(2)}^{2} \times {(2)}^{- 2} \Rightarrow x = 2^{2} \times \frac{1}{2^{2}} \Rightarrow x = 1 Now, x^{- 5} = \frac{1}{x^{5}} = \frac{1}{1^{5}} = 1$

Hence, if x = 8^2/3 × 32^–2/5, then x^–5= 1.

Page No 2.32:

Question 5:

If 6ⁿ = 1296, then 6^n–3 = _________.

Answer:

$Let 6^{n} = 1296 \Rightarrow 6^{n} = 6^{4} \Rightarrow n = 4 Now, 6^{n - 3} = 6^{4 - 3} = 6^{1} = 6$

Hence, if 6ⁿ = 1296, then 6^n–3 = 6.

Page No 2.33:

Question 6:

The value of 4 × (256)^–1/4 ÷ (243)^1/5 is ________.

Answer:

$Let x = 4 \times {(256)}^{- \frac{1}{4}} \div {(243)}^{\frac{1}{5}} \Rightarrow x = 4 \times {(4^{4})}^{- \frac{1}{4}} \div {(3^{5})}^{\frac{1}{5}} \Rightarrow x = 4 \times {(4)}^{4 \times (- \frac{1}{4})} \div {(3)}^{5 \times \frac{1}{5}} \Rightarrow x = 4 \times {(4)}^{- 1} \div {(3)}^{1} \Rightarrow x = 4 \times \frac{1}{4} \div {(3)}^{1} \Rightarrow x = 1 \div 3 \Rightarrow x = \frac{1}{3}$

Hence, the value of 4 × (256)^–1/4 ÷ (243)^1/5 is $\frac{1}{3}$ .

Page No 2.33:

Question 7:

If ${(6 x)}^{6} = 6^{2^{3}}, then x =________.$

Answer:

$Let {(6 x)}^{6} = 6^{2^{3}} \Rightarrow {(6 x)}^{6} = 6^{8} \Rightarrow 6^{6} x^{6} = 6^{8} \Rightarrow x^{6} = \frac{6^{8}}{6^{6}} \Rightarrow x^{6} = 6^{8 - 6} \Rightarrow x^{6} = 6^{2} \Rightarrow x = {(6^{2})}^{\frac{1}{6}} \Rightarrow x = 6^{2 \times \frac{1}{6}} \Rightarrow x = 6^{\frac{1}{3}}$

Hence, if ${(6 x)}^{6} = 6^{2^{3}}, then x = 6^{\frac{1}{3}} .$

Page No 2.33:

Question 8:

${\{{(\frac{a}{b})}^{\sqrt{99} - \sqrt{97}}\}}^{\sqrt{99} + \sqrt{97}}$ = ___________.

Answer:

${\{{(\frac{a}{b})}^{\sqrt{99} - \sqrt{97}}\}}^{\sqrt{99} + \sqrt{97}} = {(\frac{a}{b})}^{(\sqrt{99} - \sqrt{97}) \times (\sqrt{99} + \sqrt{97})} = {(\frac{a}{b})}^{{(\sqrt{99})}^{2} - {(\sqrt{97})}^{2}} (using the identity : a^{2} - b^{2} = (a + b) (a - b)) = {(\frac{a}{b})}^{99 - 97} = {(\frac{a}{b})}^{2} = \frac{a^{2}}{b^{2}}$

Hence, ${\{{(\frac{a}{b})}^{\sqrt{99} - \sqrt{97}}\}}^{\sqrt{99} + \sqrt{97}}$ = $\frac{a^{2}}{b^{2}} .$

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Question 9:

If $\frac{{(p + \frac{1}{q})}^{p - q} {(p - \frac{1}{q})}^{p + q}}{{(q + \frac{1}{p})}^{p - q} {(q - \frac{1}{p})}^{p + q}} = {(\frac{p}{q})}^{x}, then x =_________.$

Answer:

$\frac{{(p + \frac{1}{q})}^{p - q} {(p - \frac{1}{q})}^{p + q}}{{(q + \frac{1}{p})}^{p - q} {(q - \frac{1}{p})}^{p + q}} = {(\frac{p}{q})}^{x} \Rightarrow \frac{{(\frac{q p + 1}{q})}^{p - q} {(\frac{p q - 1}{q})}^{p + q}}{{(\frac{p q + 1}{p})}^{p - q} {(\frac{p q - 1}{p})}^{p + q}} = {(\frac{p}{q})}^{x} \Rightarrow \frac{\frac{{(q p + 1)}^{p - q}}{{(q)}^{p - q}} \times \frac{{(p q - 1)}^{p + q}}{{(q)}^{p + q}}}{\frac{{(p q + 1)}^{p - q}}{{(p)}^{p - q}} \times \frac{{(p q - 1)}^{p + q}}{{(p)}^{p + q}}} = {(\frac{p}{q})}^{x} \Rightarrow \frac{{(q p + 1)}^{p - q}}{{(q)}^{p - q}} \times \frac{{(p q - 1)}^{p + q}}{{(q)}^{p + q}} \times \frac{{(p)}^{p - q}}{{(p q + 1)}^{p - q}} \times \frac{{(p)}^{p + q}}{{(p q - 1)}^{p + q}} = {(\frac{p}{q})}^{x} \Rightarrow \frac{{(p)}^{p - q}}{{(q)}^{p - q}} \times \frac{{(p)}^{p + q}}{{(q)}^{p + q}} = {(\frac{p}{q})}^{x} \Rightarrow \frac{{(p)}^{p - q + p + q}}{{(q)}^{p - q + p + q}} = {(\frac{p}{q})}^{x} \Rightarrow \frac{{(p)}^{2 p}}{{(q)}^{2 p}} = {(\frac{p}{q})}^{x} \Rightarrow {(\frac{p}{q})}^{2 p} = {(\frac{p}{q})}^{x} \Rightarrow 2 p = x \Rightarrow x = 2 p$

Hence, if $\frac{{(p + \frac{1}{q})}^{p - q} {(p - \frac{1}{q})}^{p + q}}{{(q + \frac{1}{p})}^{p - q} {(q - \frac{1}{p})}^{p + q}} = {(\frac{p}{q})}^{x}, then x = 2 p .$

Page No 2.33:

Question 10:

If 5ⁿ⁺² = 625, then (12n + 3)^1/3 = _________.

Answer:

$Let 5^{n + 2} = 625 \Rightarrow 5^{n + 2} = 5^{4} \Rightarrow n + 2 = 4 \Rightarrow n = 2 Now, {(12 n + 3)}^{\frac{1}{3}} = {(12 (2) + 3)}^{\frac{1}{3}} = {(24 + 3)}^{\frac{1}{3}} = {(27)}^{\frac{1}{3}} = {(3^{3})}^{\frac{1}{3}} = {(3)}^{3 \times \frac{1}{3}} = 3$

Hence, if 5ⁿ⁺² = 625, then (12n + 3)^1/3 = 3.

Page No 2.33:

Question 11:

If ${(\frac{a^{3 - 7 a} \times a^{6 - 2 a}}{a^{2 a} \times a^{9 - 2 a}})}^{\frac{1}{9}} =$ __________.

Answer:

${(\frac{a^{3 - 7 a} \times a^{6 - 2 a}}{a^{2 a} \times a^{9 - 2 a}})}^{\frac{1}{9}} = {(\frac{a^{3 - 7 a + 6 - 2 a}}{a^{2 a + 9 - 2 a}})}^{\frac{1}{9}} = {(\frac{a^{9 - 9 a}}{a^{9}})}^{\frac{1}{9}} = {(a^{9 - 9 a - 9})}^{\frac{1}{9}} = {(a^{- 9 a})}^{\frac{1}{9}} = {(a)}^{- 9 a \times \frac{1}{9}} = {(a)}^{- a} = \frac{1}{a^{a}}$

Hence, ${(\frac{a^{3 - 7 a} \times a^{6 - 2 a}}{a^{2 a} \times a^{9 - 2 a}})}^{\frac{1}{9}} = \frac{1}{a^{a}} .$

Page No 2.33:

Question 12:

If $\frac{1}{{(243)}^{x}} = {(729)}^{y} = 3^{3}$ , then 5x + 6y = __________.

Answer:

$Given : \frac{1}{{(243)}^{x}} = {(729)}^{y} = 3^{3} \frac{1}{{(243)}^{x}} = 3^{3} \Rightarrow \frac{1}{{(3^{5})}^{x}} = 3^{3} \Rightarrow \frac{1}{3^{5 x}} = 3^{3} \Rightarrow 3^{- 5 x} = 3^{3} \Rightarrow - 5 x = 3 \Rightarrow 5 x = - 3 . . . (1) {(729)}^{y} = 3^{3} \Rightarrow {(3^{6})}^{y} = 3^{3} \Rightarrow 3^{6 y} = 3^{3} \Rightarrow 6 y = 3 . . . (2) Adding (1) and (2), we get 5 x + 6 y = - 3 + 3 = 0$

Hence, 5x + 6y = 0.

Page No 2.33:

Question 13:

If 6^x–y = 36 and 3^x+y= 729, then x² – y² = _________.

Answer:

$Given : 6^{x - y} = 36 and 3^{x + y} = 729 6^{x - y} = 36 \Rightarrow 6^{x - y} = 6^{2} \Rightarrow x - y = 2 . . . (1) 3^{x + y} = 729 \Rightarrow 3^{x + y} = 3^{6} \Rightarrow x + y = 6 . . . (2) Adding (1) and (2), we get 2 x = 8 \Rightarrow x = 4 Substituting the value of x in (2), we get 4 + y = 6 \Rightarrow y = 2 Now, x^{2} - y^{2} = 4^{2} - 2^{2} = 16 - 4 = 12$

Hence, x² – y² = 12.

Page No 2.33:

Question 14:

$\sqrt[4]{\sqrt[3]{2^{2}}}$ equals __________.

Answer:

$\sqrt[4]{\sqrt[3]{2^{2}}} = {\{{(2^{2})}^{\frac{1}{3}}\}}^{\frac{1}{4}} = {(2^{2})}^{\frac{1}{3} \times \frac{1}{4}} = {(2^{2})}^{\frac{1}{12}} = {(2)}^{2 \times \frac{1}{12}} = {(2)}^{\frac{1}{6}}$

Hence, $\sqrt[4]{\sqrt[3]{2^{2}}}$ equals ${(2)}^{\frac{1}{6}} .$

Page No 2.33:

Question 15:

The product $\sqrt[3]{2} . \sqrt[4]{2} . \sqrt[12]{32}$ is equal to ________.

Answer:

$\sqrt[3]{2} . \sqrt[4]{2} . \sqrt[12]{32} = {(2)}^{\frac{1}{3}} . {(2)}^{\frac{1}{4}} . {(32)}^{\frac{1}{12}} = {(2)}^{\frac{1}{3}} . {(2)}^{\frac{1}{4}} . {(2^{5})}^{\frac{1}{12}} = {(2)}^{\frac{1}{3}} . {(2)}^{\frac{1}{4}} . {(2)}^{5 \times \frac{1}{12}} = {(2)}^{\frac{1}{3}} . {(2)}^{\frac{1}{4}} . {(2)}^{\frac{5}{12}} = {(2)}^{\frac{1}{3} + \frac{1}{4} + \frac{5}{12}} = {(2)}^{\frac{4 + 3 + 5}{12}} = {(2)}^{\frac{12}{12}} = {(2)}^{1} = 2$

Hence, the product $\sqrt[3]{2} . \sqrt[4]{2} . \sqrt[12]{32}$ is equal to 2.

Page No 2.33:

Question 16:

$\sqrt[4]{{(81)}^{- 2}}$ is equal to _________.

Answer:

$\sqrt[4]{{(81)}^{- 2}} = {\{{(81)}^{- 2}\}}^{\frac{1}{4}} = {(81)}^{- 2 \times \frac{1}{4}} = {(81)}^{- \frac{1}{2}} = {(3^{4})}^{- \frac{1}{2}} = {(3)}^{4 \times (- \frac{1}{2})} = {(3)}^{- 2} = \frac{1}{3^{2}} = \frac{1}{9}$

Hence, $\sqrt[4]{{(81)}^{- 2}}$ is equal to $\frac{1}{9}$ .

Page No 2.33:

Question 17:

The value of (256)^0.16 × (256)^0.09is ________.

Answer:

${(256)}^{0.16} \times {(256)}^{0.09} = {(256)}^{0.16 + 0.09} = {(256)}^{0.25} = {(256)}^{\frac{25}{100}} = {(256)}^{\frac{1}{4}} = {(2^{8})}^{\frac{1}{4}} = {(2)}^{8 \times \frac{1}{4}} = {(2)}^{2} = 4$

Hence, the value of (256)^0.16 × (256)^0.09is 4.

Page No 2.33:

Question 1:

Write ${(625)}^{- 1 / 4}$ in decimal form.

Answer:

We have to writein decimal form. So,

${(625)}^{\frac{- 1}{4}} = \frac{1}{625^{\frac{1}{4}}} = {(\frac{1}{(5^{4})})}^{\frac{1}{4}}$

Hence the decimal form of is

Page No 2.33:

Question 2:

State the product law of exponents.

Answer:

State the product law of exponents.

If is any real number and , are positive integers, then

By definition, we have

(factor) ( factor)

to factors

Thus the exponent "product rule" tells us that, when multiplying two powers that have the same base, we can add the exponents.

Page No 2.33:

Question 3:

State the quotient law of exponents.

Answer:

The quotient rule tells us that we can divide two powers with the same base by subtracting the exponents. If a is a non-zero real number and m, n are positive integers, then

We shall divide the proof into three parts

(i) when

(ii) when

(iii) when

Case 1

When

We have

$\frac{a^{m}}{a^{n}} = \frac{a \times a \times a . . . . to m factors}{a \times a \times a . . . . t o n f a c t o r s} \frac{a^{m}}{a^{n}} = a \times a \times a . . . . t o (m - n) f a c t o r s \frac{a^{m}}{a^{n}} = a^{m - n}$

Case 2

When

We get

Cancelling common factors in numerator and denominator we get,

By definition we can write 1 as

Case 3

When

In this case, we have

Hence, whether, or,

Page No 2.33:

Question 4:

State the power law of exponents.

Answer:

The "power rule" tell us that to raise a power to a power, just multiply the exponents.

If a is any real number and m, n are positive integers, then

We have,

factors

Hence,

Page No 2.34:

Question 5:

If 2⁴ × 4² =16^x, then find the value of x.

Answer:

We have to find the value of x provided

So,

By equating the exponents we get

Hence the value of x is .

Page No 2.34:

Question 6:

If 3^x^-1 = 9 and 4^y⁺² = 64, what is the value of $\frac{x}{y}$ ?

Answer:

We have to find the value of for

So,

By equating the exponent we get

Let’s take

By equating the exponent we get

By substituting in we get

Hence the value of is

Page No 2.34:

Question 7:

Write the value of $\sqrt[3]{7} \times \sqrt[3]{49} .$

Answer:

We have to find the value of . So,

By using law rational exponents we get,

Hence the value of is

Page No 2.34:

Question 8:

Write ${(\frac{1}{9})}^{- 1 / 2} \times (64)^{- 1 / 3}$ as a rational number.

Answer:

We have to find the value of . So,

Hence the value of the value of is .

Page No 2.34:

Question 9:

Write the value of $\sqrt[3]{125 \times 27}$ .

Answer:

We have to find the value of $\sqrt[3]{125 \times 27}$ . So,

$\sqrt[3]{125 \times 27} = \sqrt[3]{5^{3} \times 3^{3}} = 5 \times 3 = 15$

Hence the value of the value of is .

Page No 2.34:

Question 10:

For any positive real number x, find the value of

${(\frac{x^{a}}{x^{b}})}^{a + b} \times {(\frac{x^{b}}{x^{c}})}^{b + c} \times {(\frac{x^{c}}{x^{a}})}^{c + a}$

Answer:

We have to find the value of L =

By using rational exponents, we get

By using rational exponents we get

By definition we can write as 1

Hence the value of expression is .

Page No 2.34:

Question 11:

Write the value of ${\{5 (8^{1 / 3} + 27^{1 / 3})^{3}\}}^{1 / 4} .$

Answer:

We have to find the value of. So,

By using rational exponents we get

Hence the simplified value of is

Page No 2.34:

Question 12:

Simplify ${[{\{{(625)}^{- 1 / 2}\}}^{- 1 / 4}]}^{2}$

Answer:

We have to simplify. So,

Hence, the value of is

Page No 2.34:

Question 13:

For any positive real number x, write the value of

${\{{(x^{a})}^{b}\}}^{\frac{1}{a b}} {\{{(x^{b})}^{c}\}}^{\frac{1}{b c}} {\{{(x^{c})}^{a}\}}^{\frac{1}{c a}}$

Answer:

We have to simplify. So,

By using rational exponents, we get

Hence the value of is

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Question 14:

If (x − 1)³ = 8, What is the value of (x + 1)² ?

Answer:

We have to find the value of , where

Consider

By equating the base, we get

By substituting in

Hence the value of is .

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