SChand CLASS 9 Chapter 6 Indices/Exponent TEST

 TEST

 Question 1

Determine whether each equation is true or false. Change the right side of the equation to make a true equation.

(i) $(2 a)^{-3}=\frac{2}{a^{3}}$

(ii) $\left(\left(a^{-1}\right)^{-1}\right)^{-1} =\frac{1}{a}$

(iii) $(2+3)^{-1}=2^{-1}+3^{-1}$

(iv) If $x \neq \frac{1}{3}$ , then $(3x – 1)^0 = (1 – 3x)^0$

Why is the condition $x \neq \frac{1}{3}$ given in part (iv) above?

Sol:

(i)$(2 a)^{-3}=\frac{2}{a^{3}}$

$(2 a)^{-3}=\frac{1}{(2 a)^{3}}=\frac{1}{8 a^{3}} \neq \frac{2}{a^{3}} \quad$ [it is not equal]

(ii) $\begin{aligned}\left(\left(a^{-1}\right)^{-1}\right)^{-1} &=\frac{1}{a} \\\left(\left(a^{-1}\right)^{-1}\right)^{-1} &=a\left(^{-1)} \times(-1) \times(-1)\right.\\ &=a^{-1}=\frac{1}{a}=\frac{1}{a} \end{aligned}$

[It is made true equation]

(iii) $(2+3)^{-1}=2^{-1}+3^{-1}$

$(2+3)^{-1}=5^{-1}=\frac{1}{5} .$

$=2^{-1}+3^{-1} .$

$=\frac{1}{2}+\frac{1}{3}=\frac{3+2}{6}$

$\begin{aligned} &=\frac{5}{6} \\ \because \frac{1}{5} & \neq \frac{5}{6} \end{aligned}$ it is not equal 

[It is not make proper equation]

(iv) $x \neq \frac{1}{3}$

$(3 x-1)^{0}=(1-3 x)^{\circ}$

$(3 x-1)^{\circ}=1$

$(1-3 x)^{0}=1 \quad\left(x^{\circ}=1\right)$

1=1 [It is make true equation]

According to Question if  $x=1 / 3$,

Then

$\left(3 x-1^{0}\right)=\left(3 \times \frac{1}{3}-1\right)^{0}$

$=(1-1)^{\circ}=0^{\circ}$

It is not make a proper equation which is not possible 

Simplify:

Question 2

$\frac{\left(5 x^{3} y^{-3} z\right)^{-2}}{y^{4} z^{-2}}$

Sol:

$\begin{aligned} & \frac{\left(5 x^{3} y^{-3} z\right)^{-2}}{y^{4} z^{-2}} \\=& \frac{\left(s^{-2}\right) x^{-6} y+1 z^{-2}}{y^{4} z^{-2}} \\=& \frac{1}{25} \frac{1}{x^{6}} \times y^{6-4} \cdot z^{-2+2} \\=& \frac{1}{25} \frac{y^{2}}{x^{6}} \times z^{0}=\frac{y^{2}}{25 x^{6}} \end{aligned}$ Ans 

 Question 3

$\left(\frac{8 a^{3} b^{-4}}{64 a^{-9} b^{2}}\right)^{2 / 3} $

Sol:

$\begin{aligned} &\left(\frac{8 a^{3} b^{-4}}{64 a^{-9} b^{2}}\right)^{2 / 3} \\=&\left[\frac{2 \times a^{3+9}}{8 \times b^{2+4}}\right]^{2 / 3} \\=&\left(\frac{a^{12}}{8 b^{6}}\right)^{2 / 3} \\=&\left(\frac{a^{12}}{2^{3} \times b^{6}}\right)^{2 / 3} \end{aligned}$

$\frac{a^{12 \times 2 / 3}}{2^{3 \times 2 / 3} \cdot b^{6 \times 2 / 3}}=\frac{a^{24 / 3}}{2^{6 / 3} \cdot b^{12 / 3}}$

$\frac{a^{8}}{2^{2} \cdot b^{4}} \Rightarrow \frac{a^{8}}{4 b^{4}}$

 Question 4

$-\sqrt[4]{16 a^{4} b^{8}}$

Sol:

$\begin{aligned}-\sqrt[4]{16 a^{4} b^{8}} &=-\left(2^{4} a^{4} b^{8}\right)^{1 / 4} \\ &=-\left(2^{a \times \frac{1}{4}} a^{4 \times \frac{1}{4}} b^{8 \times \frac{1}{4}}\right) \\ &=-\left(2^{1} a^{1} b^{2}\right) \\ &=-2 a b^{2} \text { Ans } \end{aligned}$

 Question 5

$\left[\frac{y^{2 / 3} \cdot y^{-5 / 6}}{y^{1 / 5}}\right]^{9}$

Sol:

=$\left[\frac{y^{2 / 3} \cdot y^{-5 / 6}}{y^{1 / 5}}\right]^{9}$

=$\left(y \frac{12-15-2}{18}\right)^{9}$

$=y^{-\frac{5}{2}}$

$=\frac{1}{y^{\frac{2}{5}}}$

 Question 6

$\left[\sqrt[3]{\sqrt{x^{6}}}\right.$

Sol:

$\begin{aligned} &\left[\sqrt[3]{\sqrt{x^{6}}}\right.\\=&\left[\left(x^{6}\right)^{1 / 2}\right]^{1 / 3} \\=& x^{6 \times 1 / 2} \times 1 / 3 \\=& x^{1}=x \end{aligned}$

 Question 7

Which of the following is (are) equivalent to $16\frac{–1}{2}$?

(a) – 8

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

(c) – 4

(d) $4^{-1}$

Sol:

=$16^{-1\2}$

=$\left(4^{2}\right)^{\frac{-1}{2}}$

=$4^{2 \times(-1 / 2)}$

=$4^{-1}$

$=\frac{1}{4}$ option (b) and (d) both are correct

 Question 8

Which of the following is undefined?

(a) $– 25\frac{1}{2}$

(b) $25\frac{1}{2}$

(c) $– 25\frac{–1}{2}$

(d) $(-25)\frac{1}{2}$

Sol: 

option (d) $(-25)^{1 / 2}$ is correct

because- Square root of negative number is not defined

 Question 9

True or False?

(a) $\frac{a^{4n}}{a^n}=a^4$

(b) $\frac{1}{am−n}=d^{n−m}$

(c) $a^{-n}.a^n = 1$

(d) $\frac{a^n}{b^m}=\left(\frac{a}{b}\right)^{n−m}$

Sol :

(i)

Sol: $\begin{aligned} \frac{a^{4 n}}{a^{n}} &=a^{4 n-n} \\ &=a^{3 n} \neq a^{4} \end{aligned}$

It is false 

(ii)

$\begin{aligned} \frac{1}{a^{m-n}} &=a^{-(m-n)}=a^{-m+n} \\ &=a^{n-m} \\ &=a^{n-m} \text { it is true } \end{aligned}$

(iii) $a^{-n} \cdot a^{n}=1$

$a^{-n} \cdot a^{n}=a^{-n+n}$

$=a^{\circ}=1$ $=1$ is also true

(iv) $\frac{a^{n}}{b^{m}} \neq\left(\frac{a}{b}\right)^{a-n}$ 

It is not equal so it is false

 Question 10

(i) Solve : $(- 4.8)^k = 1$

(ii) $\sqrt[3]{\sqrt{0.000064}}$ is equal to

(a) 0.0002

(b) 0.002

(c) 0.02

(d) 0.2

Sol :

(i)

$\begin{aligned}&(-4 \cdot 8)^{k}=1 \\&(-4 \cdot 8)^{k}=(-4 \cdot 8)^{0}\end{aligned}$

comparing both sides of power

k=0

(ii) $\sqrt[3]{0.000064}$

$=(0.04 \times 0.04 \times 0.04)^{1 / 2 \times 1 / 3}$

$=(0.2 \times 0.2 \times 0.2 \times 0.2 \times 0.2 \times 0.2)^{1 / 6}$

$=\left[(0.2)^{6}\right]^{1 / 6}=(0.2)^{6 \times \frac{1}{6}}$

$=(0.2)^{2}$

=0.2 Ans

option (d) is correct 

SChand CLASS 9 Chapter 6 Indices/Exponent Exercise 6(C)

  Exercise 6 C

Question 1

(i) $5^{0} \times 4^{-1}+8^{\tfrac{1}{3}}$

(ii) $\sqrt[3]{(64)^{-4}(125)^{-2}}$

(iii) $\left(9^{-3} \times 16^{3 / 2}\right)^{1 / 6}$

(iv) $\sqrt[3]{(16)^{-3 / 4} \times(125)^{-2}}$

(v) $(32)^{-2 / 5} \div(216)^{-2 / 3}$

Sol :

(i) $5^{0} \times 4^{-1}+8^{\tfrac{1}{3}}$

$=1 \times \frac{1}{4}+(2 \times 2 \times 2)^{\tfrac{1}{3}}$

$=\frac{1}{4}+\left(2^{3}\right)^{1 / 3}=\frac{1}{4}+\left(2^{3 \times \tfrac{1}{3}}\right)$

$=\frac{1}{4}+2^{1}$

$=2+\frac{1}{4}$

$\Rightarrow 2 \frac{1}{4}$

(ii) $\sqrt[3]{(64)^{-4}(125)^{-2}}$

$=\left[(64)^{-4}(125)^{-2}\right]^{1 / 3}$

$=\left[(64)^{-4 \times 1 / 3}\right]_{\times}\left[(125)^{-2 \times 1 / 3}\right]$

$=\left[(64)^{-4 / 3}\right] \times(125)^{-2 / 3}$

$=\left(4^{3}\right)^{-4 / 3} \times\left(5^{3}\right)^{-2 / 3}$

$=4^{3 \times(-4 / 3)} \times 5^{3 \times(-2 / 3)}$

$=4^{-4} \times 5^{-2}$

$=\frac{1}{4^{4} \times 5^{2}}=\frac{1}{4 \times 4 \times 4 \times 4 \times 5 \times 5}$

$=\frac{1}{6400}$

(iii) $\left(9^{-3} \times 16^{3 / 2}\right)^{1 / 6}$

=$\left[\left\{13)^{2}\right]^{-3} \times(24)^{3 / 2}\right]^{1 / 6}$

=$\left[(3)^{-6} \times(2)^{6}\right]^{1 / 6}$

=$3^{-6 \times 1 / 6} \times 2^{6} \times 1 / 6$

=$3^{-1} \times 2^{1}=\frac{1}{3} \times 2$

$=\frac{2}{3}$

(iv) $\sqrt[3]{(16)^{-3 / 4} \times(125)^{-2}}$

=$\sqrt[3]{\left(2^{4}\right)^{-3 / 4} \times\left(5^{3}\right)^{-2}}$

$=\left[2^{4 \times(-3 / 4)} \times 5^{3 \times(-2)}\right]^{1 / 3}$

=$\left[2^{-3} \times 5^{-6}\right]^{1 / 3}$

$\left[\frac{1}{2^{3} \times 5^{6}}\right]^{1 / 3}$

$\frac{1}{\left(2^{3}\right)^{1 / 2} \times\left(5^{6}\right)^{1 / 3}}=\frac{1}{2^{1} \times 5^{2}}$

$=\frac{1}{2 \times 5 \times 5}=\frac{1}{50}$

(v) $(32)^{-2 / 5} \div(216)^{-2 / 3}$

$=\left(2^{5}\right)^{-2 / 5} \div\left(6^{3}\right)^{-2 / 3}$

$=(2)^{-2 / 5} \times 5+(6)^{3 \times(-2 / 3)}$

$=2^{-2} \div 6^{-2}$

$=\left(\frac{2}{6}\right)^{-2}=\left(\frac{6}{2}\right)^{2}=\frac{6 \times 6}{2 \times 2}$

= $\frac{36}{4}$

=9

Question 2

(i) $\left(\frac{4}{125}\right)^{-2 / 3}$

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

(iii)  $\left(\frac{1}{4}\right)^{-2}-3$ $(8)^{2} / 3$ х $4^{0}+\left(\frac{3}{16}\right)^{-1 / 2}$

(iv) $16^{3 / 4}+2\left(\frac{1}{2}\right)^{-1} \times 3^0$

(v) $(81)^{3 / 4}-\left(\frac{1}{32}\right)^{-2 / 5}+(8)^{1 / 3}\left(\frac{1}{2}\right)^{-2}(2)^{0}$

Sol:

(i) $\left(\frac{4}{125}\right)^{-2 / 3}$

$\frac{1}{\left(\frac{265}{625}\right)^{1 / 4}}+\left(\frac{\sqrt{25}}{\sqrt[3]{64}}\right)^{0}$

=$\left(\frac{4}{5^{3}}\right)^{-2 / 3} \div \frac{1}{\left(\frac{4^{4}}{5^{4}}\right)^{1 / 2}}$ $+\left[\frac{\left(5^{2}\right)^{1 / 2}}{\left(4^{3}\right)^{1 / 3}}\right]^{0}$

$\left(\frac{4}{5}\right)^{3 \times(-2 / 3)} \times\left(\frac{4}{5}\right)^{4 \times 1 / 4}+1$

$=\left(\frac{4}{5}\right)^{-2} \times\left(\frac{4}{5}\right)^{1}+1$

$=\left(\frac{5}{4}\right)^{2} \times\left(\frac{5}{4}\right)^{-1}+1$

$=\left(\frac{5}{4}\right)^{2-1}+1$

$=\left(\frac{5}{4}\right)+1$

$=\frac{5}{4}+1=1 \frac{1}{4}+1$

$=2 \frac{1}{4}$ Ans

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

$=\left(3^{2}\right)^{3 / 2}-3 \times 1-\left(\frac{1}{3^{4}}\right)^{-1 / 2}$

$=3^{2} \times 3 / 2-3-\frac{1}{\left(3^{4}\right)^{-1 / 2}}$

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

$=27-3-(3)^{2}=27-3-9$

=27-12

=15 Ans

(iii)  $\left(\frac{1}{4}\right)^{-2}-3$ $(8)^{2} / 3$ х $4^{0}+\left(\frac{3}{16}\right)^{-1 / 2}$

=$\left(\frac{1}{2^{2}}\right)^{-2}-3\left(2^{3}\right)^{2 / 3} \times 1+\left[\frac{(3)^{2}}{(4)^{2}}\right]^{-1 / 2}$

$=\frac{1}{(2)^{2(-2)}}$  $-3\left(2^{3 \times 2 / 3}\right)$× 1+  $\left(\frac{3}{4}\right)^{2}$^$(-1 / 2)$

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

$=2^{4}-3 \times 4+\frac{4}{3}=16-12+\frac{4}{3}$

$=4+1 \frac{1}{3}=5 \frac{1}{3}$

(iv) $16^{3 / 4}+2\left(\frac{1}{2}\right)^{-1} \times 3$

$=\left(2^{4}\right)^{3 / 4}+2 \times 2^{1} \times 1=2^{3}+2 \times 2 \times 1$

$=2 \times 2 \times 2+2 \times 2=8+4=12$ Ans

(v) $(81)^{3 / 4}-\left(\frac{1}{32}\right)^{-2 / 5}+(8)^{1 / 3}\left(\frac{1}{2}\right)^{-2}(2)^{0}$

$=\left(3^{4}\right)^{3 / 4}-\frac{1}{2^{5(-2 / 5)}}+\left(2^{3 \times 1 / 3}\right) \times \frac{1}{(2)^{-2}} \times 1$

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

$=27-(2)^{1}+2 \times 2^{2}$

=27-4+8

=35-4

=31 Ans

Question 3

(a) Evaluate :  $x^{1 / 2} \times y^{-1} \times z^{2 / 3}$ , when x = 9, y = 2, and z = 8.

(b) Evaluate as a fraction :

$\left(\frac{27}{8}\right)^{2 / 3}-\left(\frac{1}{4}\right)^{-2}+(5)^{0}$

(c) Evaluate as a fraction :

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

Sol: 

(a) $x^{1 / 2} \times y^{-1} \times z^{2 / 3}$

x=9

y=2

z=8

$\Rightarrow(9)^{1 / 2} \times(2)^{-1} \times(8)^{2 / 3}$

$=\left(3^{2}\right)^{1 / 2} \times \frac{1}{2} \times\left(2^{3}\right)^{2 / 3}$

$=3^{2} \times 1 / 2 \times \frac{1}{2} \times 2^{3 \times 2 / 3}$

$=3^{1} \times \frac{1}{2} \times 2^{2}=3 \times \frac{1}{2} \times 4$

$=3 \times 2=6$ Ans

(b) $\left(\frac{27}{8}\right)^{2 / 3}-\left(\frac{1}{4}\right)^{-2}+(5)^{0}$

$=\left(\frac{3^{3}}{2^{3}}\right)^{2 / 3}-\left(\frac{1}{2^{2}}\right)^{-2}+1$

$\frac{3^{3 \times 2 / 3}}{2^{3 \times 2 / 3}}-\frac{1}{2^{1 \times(-2)}}+1$

$=\frac{3^{2}}{2^{2}}-\frac{1}{2^{-4}}+1=\frac{9}{4}-2^{4}+1$

$=\frac{9}{4}-16+1=2 \frac{1}{4}-16+1$

=$-12 \frac{3}{4}$ Ans.

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

= $\left(\frac{1}{2^{2}}\right)^{1 / 2}+(0.1)^{2 \times(-1 / 2)}-\left(3^{3}\right)^{2 / 3}$

$=\frac{1}{2^{2 \times 1 / 2}}+(0.1)^{-1}-3^{2}$

$=\frac{1}{2}+\frac{1}{(0.1)}-9=\frac{1}{2}+\frac{10}{1}-9$

$=10 \frac{1}{2}-9=1 \frac{1}{2}$ Ans

Question 4

(i) $\frac{(64)^{5 n / 6}(27)^{-n / 6}}{(12)^{-n / 2}} \cdot 2^{6 n}$

(ii) $\frac{6 \cdot(8)^{n+1}+162^{3 n \cdot 2}}{10 \cdot 2^{3 n+1}-7 \cdot(8)^{n}}$

(iii) $\frac{4^{2 n} \cdot 2^{n+1}}{2^{n-3} \cdot 4^{2 n+1}}$

(iv) $\frac{(5)^{2 n+3}-(25)^{n+2}}{\left((125)^{n+1}\right]^{2 / 3}}$

Sol: 

(i) $\frac{(64)^{5 n / 6}(27)^{-n / 6}}{(12)^{-n / 2}} \cdot 2^{6 n}$

=$\frac{\left(2^{6}\right)^{5 n / 6}\left(3^{9}\right)^{-n / 6}}{\left(2^{2} \times 3\right)^{-n / 2}} \cdot 2^{6 n}$

= $\frac{(2)^{6 \times \frac{5 n}{6}} \cdot(3)^{3 \times\left(\frac{-n}{6}\right)}}{(2)^{-\frac{n}{2} \times 2} \cdot(3)^{-\frac{n}{2}}} \cdot 2^{6 n}$

=$\frac{2^{5 n} \cdot 3^{-n / 2}}{2^{-n} \cdot 3^{-n / 2}} \cdot 2^{6 n}=(2)^{5 n+6 n+n} \cdot(3)^{-\frac{n}{2}}+\frac{n}{2}$

=$=2^{12 n} \cdot 3^{0}=(12)^{12 n} \cdot 1 .$

$=(2)^{12 n}$

(ii) $\frac{6 \cdot(8)^{n+1}+162^{3 n \cdot 2}}{10 \cdot 2^{3 n+1}-7 \cdot(8)^{n}}$

=$\frac{6 \cdot\left(2^{3}\right)^{n+1}+16 \cdot 2^{3 n-2}}{10 \cdot 2^{3 n+1}-7 \cdot\left(2^{3}\right)^{n}}$

= $\frac{6 \cdot 2^{3 n+3}+16 \cdot 2^{3 n-2}}{10 \cdot 2^{3 n+1}-7 \cdot 2^{3 n}}$

=$\frac{6 \cdot 2^{3 n} \cdot 2^{3}+16 \cdot 2^{3 n} \cdot 2^{-2}}{10 \cdot 2^{3 n} \cdot 2-(7 \cdot 2)^{3 n}}$

=$\frac{2^{3 n}\left(6 \times 2^{3}+16 \times 2^{-2}\right)}{2^{3 n}(10 \times 2-7)}$

=$\frac{6 \times 8+16 \times \frac{1}{4}}{20-7}$

$=\frac{48+4}{13}=\frac{52}{13}=4$

(iii) $\frac{4^{2 n} \cdot 2^{n+1}}{2^{n-3} \cdot 4^{2 n+1}}$

$=\frac{\left(2^{2}\right)^{2 n} \cdot 2^{n+1}}{2^{(n-3)}\left(2^{2}\right)^{2 n-1}}$

=$=\frac{2^{4 n} \cdot 2^{n+1}}{2^{n-3} \cdot 2^{4 n+2}}=\frac{2^{4 n+n+1}}{2^{n-3+4 n+2}}$

=$\frac{25 n+1}{25 n-1}$

$=25 n+1-5 n+1$

$=2^{2}$

=4

(iv) $\frac{(5)^{2 n+3}-(25)^{n+2}}{\left((125)^{n+1}\right]^{2 / 3}}$

=$\frac{5^{3 n+3}-(5)^{m+2}}{\left[\left(5^{3}\right)^{n+3}\right]^{2 / 3}}$

=$\frac{5^{2 n} \cdot 5^{3} \cdot 5^{2 n} \cdot 5^{4}}{5(3 n+5)^{3 / 2}}$

$\Rightarrow \frac{5^{2 n} \cdot 5^{3}-5^{2 n} \cdot 5^{4}}{5^{2 n} \cdot 5^{2}}$

=$\frac{5^{2 n}\left(5^{3}-5^{4}\right)}{5^{2} n \cdot 5^{2}}$

=$\frac{125-625}{25}$

$=\frac{-500}{25}$

=-20 Ans

Question 5

If  $\frac{49^{n+1} \cdot 7^n -(343)^n}{7^{3m} \cdot 2^n}=\frac{3}{343}$  , prove that m = (m + 1)

Sol:

$\begin{aligned} & \frac{49^{n+1} \cdot 7^{n}-(343)^{n}}{7^{3 m} \cdot 2^{n}} \Rightarrow \frac{3}{343} \\ \Rightarrow & \frac{\left(7^{2}\right)^{n+1} \cdot 7^{n}-\left(7^{3}\right)^{n}}{7^{3 m} \cdot 2^{n}}=\frac{3}{343} \\ \Rightarrow & \frac{7^{n}+2 \cdot 7^{n}-7^{3 n}}{7^{3 m} \cdot 2^{4}}-\frac{3}{343} \\ \Rightarrow & \frac{7^{3 n}\left(7^{2}-1\right)}{16 \cdot 7^{3 m}}=\frac{3}{343} \\ \Rightarrow & \frac{7^{3 n} \times 48}{16 \cdot 7^{3 m}}=\frac{3}{343} \end{aligned}$

=$\frac{7^{3 n}}{7^{3 m}}=\frac{3 \times 16}{343 \times 48}=\frac{1}{343}$

=$\frac{\left(7^{n}\right)^{3}}{\left(7^{m}\right)^{3}}=\frac{1}{7^{3}}=\left(\frac{1}{7}\right)^{3} \quad \Rightarrow \frac{7^{n}}{7^{m}}=\frac{1}{7} 7.7 n=7^{m}$

$\Rightarrow \quad 7^{n+1}=7^{m}$ (comparing both sides) $\Rightarrow m=n+1$ 

Hence proved.

Question 6

If $2205=3^{a} \times 5^{b} \times 7^{c}$

(i) Find the numerical values of a, b and c

(ii) Hence evaluate $3^{a} \times 5^{-b} \times 7^{-c}$

Sol:

(i) $2205=3^{a} \times 5^{b} \times 7^{c}$

$\begin{array}{l|l}3 & 2205 \\\hline 3 & 735 \\\hline 5 & 245 \\\hline 7 & 49 \\\hline 7 & 7 \\\hline 1\end{array}$

On comparing = a=2

b=1

c=2 

so  $3^{2} \times 5^{1} \times 7^{2}=3^{a} \times 5^{b} \times 7^{c}$

(ii) 

$\begin{aligned} & 3^{a} \times 5^{-b} \times 7^{-c} \\=& \frac{3^{a}}{5^{b} \times 7^{c}}=\frac{3^{2}}{5^{1} \times 7^{2}} \\=& \frac{9}{5 \times 49} \Rightarrow \frac{9}{245} \\ & \frac{9}{245} \text { Ans. } \end{aligned}$

Question 7

If $\left[\frac{b^3c^{-2}}{b^{-4}c^3}\right]^{-3}+\left[\frac{b^{-1}c}{b^2.c^{-2}}\right]^5=b^x.c^x$ , prove that x + y + 6 = 0

Sol:

$\left[\frac{b^{3} c^{-2}}{b^{-4} c^{3}}\right]^{-3} \div\left[\frac{b^{-1} c}{b^{2} \cdot c^{-2}}\right]^{5}=b^{x} \cdot c^{y}$

Now $\text { R.H.S. }=x+y+6=-6+0+6=0$

Hence proved.

Question 8

Prove that

(i) $(x+y)^{-1}\left(x^{-1}+y-1\right)=\frac{1}{x y}$

(ii) $\left(x^{-1}+y-1\right)^{-1}=\frac{x y}{x+y}$

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

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

(v) $a b \sqrt{\frac{x^{a}}{x^{b}}} \cdot b c \sqrt{\frac{x^{b}}{x^{c}}} \cdot \sqrt[ca]{\frac{x^{2}}{x^{a}}}=1$

Sol:

(i) 

$\begin{aligned}(x&+y)^{-1}\left(x^{-1}+y-1\right)=\frac{1}{x y} \\ \text { L.H.S. } &=(x+y)^{-1}\left(x^{-1}+y-2\right) \\ &=\frac{1}{x+y} \times\left(\frac{1}{x}+\frac{1}{y}\right) \\ &=\frac{1}{x+y} \times \frac{y+5}{x y} \end{aligned}$

$=\frac{1}{x+y} \times \frac{x+y}{x y} \Rightarrow \frac{1}{x y}$

Hence pruerd,

$\text { L.H.S }=\text { R.H.S. }$

(ii) $\left(x^{-1}+y-1\right)^{-1}=\frac{x y}{x+y}$

L.H.S 

$\begin{aligned} &\left(x^{-1}+y^{-1}\right)^{-1} \\=&\left[\frac{1}{x}+\frac{1}{y}\right]^{-1} \\=&\left[\frac{y+x}{x y}\right]^{-1}=\frac{x y}{x+y} \quad \text { R.H.S. } \end{aligned}$

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

L.H.S

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

$=\frac{a+b+c}{\frac{c+a+b}{a b c}} \Rightarrow(a+b+c) \times \frac{a b c}{(a+b+c)}$

=$a b c$ R.H.S (Hence proved)

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

 

L.H.S

$=\frac{1}{x^{a-a}+x^{b-a} \cdot x^{c-a}}+\frac{1}{x^{b-b}+x^{a-b}+x^{c-b}}+\frac{1}{x^{c-c}+x^{b-c}+x^{a-c}}$

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

$=\frac{1}{\left(x^{a}+x^{b}+x^{c}\right)}\left[\frac{1}{x^{-a}}+\frac{1}{x^{-b}}+\frac{1}{x-c}\right]$

$\frac{1}{\left(x^{a}+-x^{b}+x^{2}\right)}\left[\frac{x^{a}+x^{b}+x^{c}}{1}\right]=1$

L.H.S =R.H.S PROVED

(v) $a b \sqrt{\frac{x^{a}}{x^{b}}} \cdot b c \sqrt{\frac{x^{b}}{x^{c}}} \cdot \sqrt[ca]{\frac{x^{2}}{x^{a}}}=1$

$x^{\frac{a-b}{a b}} \cdot x \frac{b-c}{b c} \cdot x^{\frac{c-a}{c a}}$

$=x \frac{a-b}{a b}+\frac{b-c}{b c}+\frac{c-a}{c a}$

$x \frac{a b c-b c + a b-a c+b c-a b}{a b c}$

$x \frac{D}{a b c}$

$\begin{aligned}&=x^{0}=1 \\&L \cdot H \cdot S=R \cdot H \cdot S\end{aligned}$

Hence, proved

Question 9

(i) $\left(p^{1 / 3}-p^{-1/ 3}\right)\left(p^{2 / 3}+1+p^{-2 / 3}\right)$

(ii) $(\sqrt{11}+\sqrt{3})^{1 / 3}(\sqrt{11}-\sqrt{3})^{1 / 3}$

Sol: 

(i)$\left(p^{1 / 3}-p^{-1/ 3}\right)\left(p^{2 / 3}+1+p^{-2 / 3}\right)$

if $p^{1 / 3}=x$ and $p^{-1 / 3}=y$

$=\left(a^{2}-y^{2}\right)\left(x^{2}+x y+y^{2}\right)$

$=x^{3}-y^{3}=\left(p^{1 / 3}\right)^{3}-\left(p^{-1 / 3}\right)^{3}$ $=p^{1}-p^{-1}=p-\frac{1}{p}$ Ans.

(ii) $(\sqrt{11}+\sqrt{3})^{1 / 3}(\sqrt{11}-\sqrt{3})^{1 / 3}$

if $\sqrt{11}=x$

$\sqrt{3}=y$

$=(x+y)^{1 / 3}(x-y)^{1 / 3}$

$[$ from formula. $]$= $\left(x^{2}-y^{2}\right)^{1 / 3}$

$=\left[(\sqrt{11})^{2}-(\sqrt{3})^{2}\right]^{1 / 3}$

$=(11-3)^{1 / 3}=(8)^{1 / 3}$

$=\left(2^{3}\right)^{1 / 3}=2^{3 / \frac{1}{3}}$

$=2^{2}=2$ Ans

Question 10

If $3^{x}=5^{y}=45^{z}$ , prove that $x =\frac{2yz}{y-z}$

Sol: 

L.H.S 

Suppose $3^{x}=5^{y}=45^{z}=a$ 

so, $\begin{aligned} a^{1 / x} &=3, a^{1 / y}=5, \frac{1}{a^{2}}=45 \\ & 45=3 \times 3 \times 5 \\ & \Rightarrow 3^{2} \times 5=45 \\ &=a^{\frac{1}{z}}=a^{\frac{2}{x}} \times a^{1 / y} \\ &=a^{\frac{1}{z}}=a^{\frac{2}{x}+\frac{1}{y}} \end{aligned}$

In comparing

$\quad \frac{1}{2}=\frac{2}{x}+\frac{1}{y}=\frac{2}{x}=\frac{1}{z}-\frac{1}{y}$ $\Rightarrow \quad \frac{2}{x}=\frac{y-z}{y z}$ $\Rightarrow \frac{1}{x}=\frac{y-z}{2 y z}$ $\Rightarrow \quad x=\frac{2 y z}{y-z}$ R.H.S. 

Hence proved

Question 11

If $x = ab^{p-1}, y = ab^{q-1}$ and $z = ab^{r-1}$ prove that $x^{p-r}x y^{r-p}.z^{p-q} = 1$.

Sol;

L.H.S.

$\begin{aligned}&x=a b^{p-1} \\&y=a b^{q-1} \\&z=a b^{r-1}\end{aligned}$

So $=x^{-r-r} \cdot y^{r-p} \cdot z^{p-q}$

$=\left(a b^{p-1}\right)^{q-\dot{r}} \cdot\left(a b^{2-1}\right)^{r-p} \cdot\left(a b^{r-1}\right)^{p-q}$

$=a^{q-r} \cdot b^{(2-r)(p-1)} \cdot a^{r-p} \cdot b^{(r-p)(r-1)} \cdot a^{p-2} \cdot b^{(p-2)}(r-1)$

$=a^{\circ} \cdot b^{\circ}$

$=1 \times 1=1$ R.H.S 

Hence proved

SChand CLASS 9 Chapter 6 Indices/Exponent Exercise 6(B)

 Exercise 6(B)

 Question 1

Evaluate:

(i) 5°

(ii) $2^{-3}$

(iii) $10^{-4}$

(iv) $(- 2)^{-2}$

(v) $\left(\frac{2}{3}\right)^{\circ}$

(vi) $\left(\frac{3}{4}\right)^{-3}$

(vii) $\left(\frac{−1}{2}\right)^{-2}$

(viii) $\frac{2^{−2}}{5^{−2}}$

(ix) $\frac{3x^0−1}{3x^0+1}$

Sol :

(i) $5^{\circ}$

= $5^{\circ}$=1   $\left(\because x^{0}=1\right)$

(ii) $2^{-3}$

$2^{1} \times 2^{-1} \times 2^{-1}$

$8^{-1}$

$\frac{1}{8}$

(iii) $10^{-4}$

$10^{-1} \times 10^{-1} \times 10^{-1} \times 10^{-1}$

$10000^{-1}$

$\frac{1}{10000}$

(iv) $(-2)^{-2}$

$\frac{1}{(-2)^{2}}$

$\frac{1}{-2 x-2}$

$\frac{1}{4}$

(v) $\left(\frac{2}{3}\right)^{0}$

=1

(vi) $\left(\frac{3}{4}\right)^{-3}$

$\left(\frac{4}{3}\right)^{3} \Rightarrow \frac{4 \times 4 \times 4}{3 \times 3 \times 3} \Rightarrow \frac{64}{27}$

(vii) $\left(-\frac{1}{2}\right)^{-2}$

$\left(-\frac{2}{1}\right)^{2}$

$(-2)^{2}$

$-2 x-2 \Rightarrow 4$

(viii) $\frac{2^{-2}}{5-2}$

$\frac{5^{2}}{2^{2}}$

$\frac{5 \times 5}{2 \times 2} \Rightarrow \frac{25}{4}$

(ix)  $\frac{3 x^{0}-1}{3 x^{0}+1}$

$\frac{3 x 1-1}{3 \times 1+1}$

$\frac{3-1}{3+1}$

$\frac{2}{4} \Rightarrow \frac{1}{2}$

Question  2

Evaluate:

(i) $4\frac{1}{2}$

(ii) $8\frac{1}{3}$

(iii) $16\frac{1}{4}$

(iv) $-27\frac{1}{3}$

(v) $32\frac{3}{5}$

(vi) $(125)^\frac{−2}{3}$

(vii) $\left(\frac{8}{125}\right)^\frac{1}{3}$

(viii) $\left(\frac{1}{216}\right)^{\frac{−2}{3}}$

(ix) $-(- 27)^\frac{−4}{3}$

(x) $\left(\frac{27}{8}\right)^\frac{−2}{3}$

(xi) $(0.0625)^{\frac{3}{4}}$

(xii) $\left(12\frac{19}{27}\right)^\frac{1}{3}$

Sol:

(i) $4 \frac{1}{2}$

$(2 \times 2)^{\frac{1}{4}}$

$\left(2^{2}\right)^{\frac{1}{2}}$

$2\left(2 \times \frac{1}{2}\right)$

$2^{1} \Rightarrow 2$

(ii)$8 \frac{1}{3}$

$(2 \times 2 \times 2)^{\frac{1}{3}}$

$2^{3 \times \frac{1}{3}}$

=21=2

(iii) $16^{\frac{1}{4}}$

$(2 \times 2 \times 2 . \times 2)^{\frac{1}{4}}$

$\left(2^{4}\right)^{\frac{1}{4}}$

=21=2

(iv)  $(-27)^{\frac{1}{3}}$

$[(-3) \times(-3) \times(-3))^{\frac{1}{3}}$

$(-3)^{3 \times \frac{1}{3}}$

$(-3)^{1}$

$(-3)^{1} \Rightarrow-3$

(v) $32^{\frac{3}{5}}$

$(2 \times 2 \times 2 \times 2 \times 2)^{\frac{3}{5}}$

$\left(2^{5}\right)^{\frac{2}{5}} \Rightarrow 2^{5 \times \frac{3}{5}}=2^{3}$

$2 \times 2 \times 2=8$

(vi) $(125)^{-3}$

$(5 \times 5 \times 5)^{-2}$

$\left(5^{3}\right)^{\frac{-2}{3}}$

$5^{3 \times \frac{-2}{3}}$

$\Rightarrow 5^{-2}$

$\frac{1}{52} \Rightarrow \frac{1}{5 \times 5} \Rightarrow \frac{1}{25}$

(vii) $\left(\frac{8}{125}\right)^{1 / 3}$

$\left(\frac{2 \times 2 \times 2}{5 \times 5 \times 5}\right)^{1 / 3}$

$\frac{2^{3 \times \frac{1}{3}}}{5^{3 \times \frac{1}{3}}} \Rightarrow \frac{2}{5}$

(viii)$\left(\frac{1}{216}\right)^{-\frac{2}{3}}$

$\left(\frac{1}{6 \times 6 \times 6}\right)^{-\frac{2}{3}}$

=$\left(\frac{1}{(6)^{3}}\right)^{\frac{-2}{3}}$

=$\frac{(1)^{\frac{-2}{3}}}{\left(6^{3}\right)^{\frac{-2}{3}}}$

=$\frac{1}{(6)^{-2}}$

=$(6)^{2}$

=36

(ix) $-(-27)^{-4\3}$

=$\Rightarrow-[(-3)(-3)(-3)]^{-\frac{4}{3}}$

=$-\left(\left(-3^{3}\right)^{-\frac{4}{3}}\right)$

=$\left[(-3)^{3 x}\left(-\frac{-4}{3}\right)\right]$

=$-(-3)^{-4}$

=$\frac{-1}{(-3)^{4}}$

=$\frac{-1}{(-3) \times(-3) \times(-3) \times(-3)}$

= $\frac{-1}{81}$

(x) $\left[\frac{27}{8}\right]^{-\frac{2}{3}}$

$\begin{aligned} &\left.\frac{3 \times 3 \times 3}{2 \times 2 \times 2}\right)^{-\frac{2}{3}} \\ \Rightarrow &\left(\frac{3^{3}}{2^{3}}\right)^{-\frac{2}{3}} \\ \Rightarrow & \frac{3^{3 x}-\frac{2}{3}}{2^{2 \times-\frac{2}{3}}} \\ \Rightarrow & \frac{3^{-2}}{2^{-2}} \\ \Rightarrow & \frac{2^{2}}{3^{2}} \\ \Rightarrow & \frac{2 \times 2}{3 \times 3} \\ \Rightarrow & \frac{4}{9} \end{aligned}$

(xi)  $(0.062 .5)^{3 / 4}$

$\left(0.5 \times 0.5 \times 0.5 \times 0.5)^{\frac{3}{4}}\right.$

$\left((0.5)^{4}\right)^{\frac{-3}{4}}$

$(0.5)^{4 \times 3}$

$(0.5)^{3} \Rightarrow 0.5 \times 0.5 \times 0.5 \Rightarrow 0.125$

(xii) $\left(12 \frac{19}{27}\right)^{\frac{1}{3}}$

=$\left(\frac{12 \times 27+9}{27}\right)^{1 / 3}$

=$\left(\frac{324+19}{27}\right)^{1 / 3}$

=$\left(\frac{343}{27}\right)^{2 / 3}$

=$\left(\frac{7 \times 7 \times 7}{3 \times 3 \times 3}\right)^{1 / 3}$

=$\left(\frac{(7)^{3}}{(3)^{3}}\right)^{1 / 3}$

=$\frac{7^{3 \times \frac{1}{3}}}{3^{8 \times \frac{1}{3}}}$

=$\frac{7}{3}$

=$2 \frac{1}{3}$

SChand CLASS 9 Chapter 6 Indices/Exponent Exercise 6(A)

 Exercise 6(A)

Question 1

Write the products in the exponential form:

(i) (x . x . x . x. x) (x. x)

(ii) -1 (n . n. n) (n . n)

(iii) $y^8 \times y^5 \times y^2$

(iv) $– x (- x^4)$

(v) $(3a^7b^8c^9)\times (5a^{27}b^{16}c^8)$

(vi) $(x + 2)^2 . (x + 2)^4$

(vii) $(2m – 32n)^{3a-2b} \times  (2m – 3n)^{6a+10b}$

(viii) $x^{2a+b-c}.x^{2c+a-b}.x^{2b+c-a}$

Sol :

(i) (x.x.x.x.x)(x.x)

$x^{5} \cdot x^{2}$

$=x^{m} \cdot x^{n} \Rightarrow x^{m+n}$

$=\quad x^{5+2} \Rightarrow x^{-2}$

(ii) $-1 .(n \cdot n, n)(n \cdot n)$

$-1 n^{3} \cdot n^{2}$

$\because x^{m} \cdot x^{n}=x^{m+n}$

$-1 n^{3+2}$

$\Rightarrow-n^{5}$

(iii) $y^{8} \times y^{5} \times y^{2}$

$\Rightarrow y^{8+5+2}$

$\Rightarrow y^{15}$

(iv) $-x\left(-x^{5}\right)$

$\Rightarrow x \times x^{5}$

$\Rightarrow x^{1+5}$

$\Rightarrow x^{6}$

(v) $\left(3 a^{7} b^{8} c^{9}\right) \times 5\left(a^{27} b^{16} c^{8}\right)$

$3 \times a^{7} \times b^{8} \times c^{9} \times 5 \times c^{27} \times b^{16} \times c^{8}$

∵$x^{m} \cdot x^{n}=x^{m+n}$

$3 \times 5\left(a^{7} \times a^{27}\right)\left(b^{8} \times b^{16}\right)\left(c^{9} \times c^{8}\right)$

$3 \times 5 a^{7+27} \times b^{8+16} c^{9+8}$

$15 a^{34} b^{24} c^{17}$

(vi) $(2 m+3 n)^{3 n-2 b} \times(2 m-3 n)^{6 a+10 b}$

$\because x^{m} \cdot x^{n} \Rightarrow x^{m+n}$

$(2 m-3 n)^{3 a-2 b+6 a+10 b}$

$(2 m-3 n)^{(3 a+6 a-2 b+10 b)}$

$(2 m-3 n)^{9 a+0 b}$

(vii) $x^{2 a+b-c} \cdot x^{2 c+a-b} \cdot x^{2 b+c-a}$

∵$x^{m} x^{n} \cdot x^{0} \Rightarrow x^{m+n+0}$

$\Rightarrow x^{2 a+b-c+2c+a-b+2b+c-a}$

$\Rightarrow x^{2 a+2c+2 b}$

Question 2

Write each expression in the simpler form:

(i) xy³

(ii) $(-x)^5$

(iii) $(-2xy)^4$

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

(v) $(x^2)^5$

(vi) $(7^3)^8$

(vii) $(6a^2)^3$

(viii) $(-x^2y^3)^3$

(ix) $(p^2)^5 \times (p^3)^2$

(x) $3(x^4y^3)^{10} \times 5(x^2y^2)^3$

(xi) $\left(\frac{c^3}{d^2}\right)^7$

(xii) $\left(\frac{3p^2}{4q^2}\right)^n$

(xiii) $\left(\frac{a^2b^2}{x^2y^3}\right)^m$

Sol :

(i) $(xy)^{3}$

∵$\left(xy\right)^{m}=x^{m} \cdot y^{m}$

$(x y)^{3}=x^{3} \cdot y^{3}$

(ii) $(-x)^{5}$

⇒(-x)×(-x)×(-x)×(-x)×(-x)

⇒(-x)⁵

(iii) $(-2 x y)^{4}$

⇒(-2xy)×(-2xy)×(-2xy)×(-2xy)

⇒(-2xy)⁴

⇒(16x⁴y⁴)

(iv) $\left(\frac{p}{q}\right)^{8}$

⇒$\frac{p^{8}}{q^{8}}$

(v) $\left(x^{2}\right)^{5}$

$=x^{2 \times 5}=x^{10}$

(vi) $\left.7^{3}\right)^{8}$

⇒$7^{3 \times 8}$

⇒$7^{24}$

(vii) $\left(6 a^{2}\right)^{3}$

⇒$\left(6 a^{2}\right) \times\left(6a^{2}\right) \times\left(6a^{2}\right)$

⇒$\left(216a^{2+2+2}\right)$

⇒$216a^6$

(viii) $\left(-x^{2} y^{3}\right)^{3}$

⇒$\left(-x^{2} y^{3}\right) \times\left(-x^{2} y^{3}\right) \times\left(-x^{2} y^{3}\right)$

⇒$\left(-x^{2+2+2} \cdot y^{3+3+3}\right)$

⇒$-x^{6} y^{6}$

(ix) $\left(p^{2}\right)^{5} \times\left(p^{3}\right)^{2}$

⇒$p^{2 \times 5+3 \times 2}$

⇒$p^{10+6}$

⇒$p^{16}$

(x) $3\left(x^{4} y^{3}\right)^{10} \times 5\left(x^{2} y^{2}\right)^{3}$

⇒$3\left(x^{4} y^{10}\cdot y^{3 \times 10}\right) \times 5\left(x^{2 \times 3} \cdot y^{2 \times 3}\right)$

⇒$3 \times 5\left(x^{40} \cdot y^{30} \cdot x^{6} \cdot y^{6}\right)$

⇒$15\left(x^{40+6} y^{30+6}\right)$

⇒$15 \cdot\left(x^{46} y^{36}\right)$

⇒$15 x^{46} y^{30}$

(xi) $\left(\frac{c^{3}}{d^{2}}\right)^{7}$

⇒$\frac{\left(c^{3}\right)^{7}}{\left(d^{2}\right)^{7}}$

⇒$\frac{c^{21}}{d ^{14}}$

(xii) $\left(\frac{3p^2}{4q^2}\right)^a$

$=\frac{3^{a} \cdot p^{2 a}}{4^{a} q^{2a}}$

(xiii) $\left(\frac{a^{2} b^{2}}{x^{2} y^{3}}\right)^{m}$

$=\frac{\left(a^{2} b^{2}\right)^{m}}{\left(x^{2} y^{3}\right)^{m}}$

$= \frac{a^{2 m} \times b^{2 m}}{x^{2 m} \times y^{3 m}}$

Question 3

Find the quotient:

(i) $x^6 + x^2$

(ii) $x^{2a} + x^4$

(iii) $\frac{p^5q^3}{p^3q^2}$

(iv) $(−8x^{27}y^{21})÷(−16x^6y^{17})$

(v) $(−8x^{27}y^{21})÷(−16x^6y^{17})$

(vi) $\frac{4pq^2(−5pq^3)}{10p^2q^2}$

(vii) $\frac{(−4ab^2)^2}{16ab}$

(viii) $\frac{x^{a−b}y^{c−d}}{x^{2b−a}y^c}$

(ix) $\frac{(m^{3n−9})^6}{m^{2n−4}}$

(x) $\left[\frac{(x^{2a−4})^2}{x^{a+5}}\right]^3$

Sol :

(i) $x^{6} \div x^{2}$

∵$x^{m} \div x^{n} \Rightarrow x^{m-n}$

$=x^{6} \div x^{2}$

$=x^{6-2}=x^{4}$

(ii) $x^{2} \div x^a$

$=x^{2a-a}$

$=x^{a}$

(iii) $\frac{p^{5} q^{3}}{p^{3} q^{2}}$

$=p^{5} q^{3} \div p^{3} q^{2}$

$=p^{5-2} q^{3-2}$

$=p^{2} q^1$

$=p^{2} q$

(iv) $\frac{-35 x^{10} y^{5}}{-7 x^{3} y^{3}}$

$=-35 x^{10} y^{5} \div\left(-7 x^{3} y^{3}\right)$

$=\frac{35^{5}}{7} x^{10-3} \cdot y^{5-3}$

$=5 x^{7} \cdot y^{2}$

(v) $-8 x^{27} y^{21} \div\left(-16 x^{6} y^{17}\right)$

$=\frac{-8}{-162} x^{27-6} y^{21-17}$

$=\frac{1}{2} x^{21} y^{4}$

(vi) $\frac{4 p q^{2}\left(-5 p q^{3}\right)}{10 p^{2} q^{2}}$

$=\frac{4 \times -5 .p^{1+1}.q^{2+3}}{10p^2q^2}$

$=\frac{-20p^2q^5}{10p^2q^2}$

$=-2 p^{2-2} q^{5-2}=-2p^0q^2$

$=-2q^2$

(vii) $\frac{\left(-4 a b^{2}\right)^{2}}{16 a b}$

$=\frac{16 a^{2} b^{2+2}}{16 a b}$

$=a^{2}.b^{4} \div a b=a^{2-1}.b^{4-1}$

$=a b^{3}$

(viii) $\frac{x^{a-b} y^{c-d}}{x^{2 b-a} y^{c}}$

$=x^{(a-b)(2 b-a)} y^{c-d-c}$

$=x^{a-b-2 b+a} y^{-d}$

$=x^{2 a-3 b} y^{-d}$

(ix) $\frac{\left(m^{3 n-9}\right)^{6}}{m^{2 n-4}}$

$=\frac{m^{3 n \times 6-9 \times 6}}{m^{2 n-4}}$

$=\frac{m^{18 n-54}}{m^{2 n-4}}$

$=m^{10 n-54-(2 n-4)}$

$=m^{18 n-54-2 n+4}$

$=m^{16n-50}$

(x) $\left[\frac{\left(x^{2a-4}\right)^{2}}{x^{a+5}}\right]^{3}$

$\Rightarrow\left(\frac{x^{2 \times 2 a-4 \times 2}}{x^{9+5}}\right)^{3}$

$\Rightarrow\left(\frac{x^{4 a-8}}{x^{a+5}}\right)^{3}$

$\Rightarrow\left(x^{4 a-8-(a+5)}\right)^{3}$

$\Rightarrow\left(x^{4 a-8-a-5}\right)^{3}$

$\Rightarrow\left(x^{3 a-13}\right)^{3}$

$\Rightarrow 3^{9a-39}$

SChand CLASS 9 Chapter 6 Indices/Exponent Exercise 6(D)

 Exercise 6 D

Question 1

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

Sol:

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

$=2^{x+1}=\left(2^{2}\right)^{x-3}$

$=2^{x+1}=2^{2 x-6}$

Comparing both sides of powers

$\begin{aligned}& x+1=2 x-6 \\\Rightarrow & 2 x-x=1+6 \\=& x=7\end{aligned}$

So= x=7 Ans 

Question 2

$ x^{\frac{-3}{4}}=\frac{1}{8} $

Sol:

$\begin{aligned} x^{-3 / 4} &=\frac{1}{8} \\ &=\frac{1}{8}=\frac{1}{2^{3}}=2^{-3} \end{aligned}$

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

$=\left(x^{1 / 4}\right)^{-3}=2^{-3}$

=$\left(x^{1 / 4}\right)=2$  doing comparing.

$=x=2^{4}$

$=x=16$

So= x=16 Ans 

Question 3

$(x-1)^{\frac{2}{3}}=25$

Sol:

$\begin{aligned}(x-1)^{2 / 3} &=25 \\ &=25=5^{2} \\(x-1)^{2 / 3} &=\left(5^{3}\right)^{2 / 3} \\ x-1 &=5^{3} \text { doing comparing  } \\ x-1 &=125 \end{aligned}$

x-1=125

x=125+1

x=126 Ans.

Question 4

$2^{5 x+3}=8^{x+3}$

Sol:

$2^{5 x+3}=8^{x+3}$

$=\left(2^{3}\right)^{x+3}$

$=2^{5 x+3}=2^{3 x+9}$

$5x+3=3x+3$ 

$5x-3x=9-3$

2x=6

x=3

Question 5

$\left(\sqrt{\frac{5}{7}}\right)^{x-1} &=\left(\frac{125}{343}\right)^{-1}$

Sol:

$\begin{aligned}\left(\sqrt{\frac{5}{7}}\right)^{x-1} &=\left(\frac{125}{343}\right)^{-1} \\ \Rightarrow\left(\frac{5}{7}\right)^{\frac{x-1}{2}} &=\left(\frac{5^{3}}{7^{3}}\right)^{-1}=\left(\frac{5}{7}\right)^{-3} \end{aligned}$

if comparing both sides power

$\begin{aligned} & \frac{x-1}{2}=-3 \\=& x-1=-6 \\ \Rightarrow & x=-6+1 \\=& x=-5 \\ & x_{0}, x=-5 \text { Ans. } \end{aligned}$

Question 6

$11^{3-4 x}=\left(\sqrt{\frac{1}{121}}\right)^{-2}

Sol:

=$11^{3-4 x}=\left(\sqrt{\frac{1}{121}}\right)^{-2}=\left(\sqrt{\frac{1}{(11)^{2}}}\right)^{-2}$

=$11^{3-4 x}=11 \frac{(-2) \times(-2)}{2}$

=$11^{3-4 x}=11^{2}$

Comparing both side of power

$\begin{aligned}3-4 x &=2 \\\Rightarrow-4 x &=2-3 \quad=-1\end{aligned}$

$\begin{aligned}&-4 x=-1 \\&x=\left(+\frac{1}{4}\right)\end{aligned}$

So $x=\frac{1}{4}$ Arus.

Question 7

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

Sol :

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

$\left(\sqrt[3]{(2)^{2}}\right)^{2 x+\frac{1}{2}}=\frac{1}{2^{5}}$ $\left\{\therefore 2^{5}=2 \times 2 \times 2 \times 2 \pi 2=32\right\}$

$\Rightarrow \quad\left[2^{3 / 3}\right]^{2 x+1 / 2}=2^{5}$

$=2^{2 / 3\left(2 x+\frac{1}{2}\right)}=2^{-5}$

If comparing 

$\frac{2}{3}\left(2 x+\frac{1}{2}\right)=-5$

$\frac{4}{3} x+\frac{1}{3}=-5$

So $\begin{aligned} & 4 x+1=-15 \\ & 4 x=-13-1=-16 \\ \therefore & x=\frac{-16}{4} \\ \therefore & x=-4 \text { Ans } \end{aligned}$

Question 8

Find the value of .Y if  $\sqrt{\frac{p}{q}}=\left(\frac{q}{p}\right)^{1-2 x}$

Sol:

$\begin{aligned} & \sqrt{\frac{p}{q}}=\left(\frac{q}{p}\right)^{1-2 x} \\ &\left(\frac{p}{q}\right)^{1 / 2}=\left(\frac{p}{2}\right)^{-(1-2 x)} \\=&\left(\frac{p}{q}\right)^{2 x-1} \\ & 2 x-1=\frac{1}{2} \\ & 2 x=\frac{1}{2}+1=\frac{3}{2} \end{aligned}$

$\Rightarrow x=\frac{3}{2 \times 2}=\frac{3}{4}$

$\therefore x=\frac{3}{4}$ Ans

Question 9

Solve for x, $2^{3}\left(5^{\circ}+3^{2 x}\right) &=8 \frac{8}{27}$

Sol:

$\begin{aligned} 2^{3}\left(5^{\circ}+3^{2 x}\right) &=8 \frac{8}{27} \quad \therefore[2778+8] \\ 8\left(1+3^{2 x}\right) &=\frac{224}{27} \quad\left[5^{\circ}=1\right] \end{aligned}$

=$1+3^{2 x}=\frac{224}{27} \times \frac{1}{-8}$

$=\frac{28}{27}$

=$3^{2 x}=\frac{28}{27}-1$

=$\frac{28-27}{27}=\frac{1}{27}$

$3^{2 x}=\frac{1}{3^{2}}=3^{-3}$

$2 x=-3$

$x=-3 / 2 \quad So \quad x=\frac{-3}{2}$ Ans

Question 10

Solve for x, $\sqrt{\left(8^{0}+\frac{2}{3}\right)}=(0.6)^{2-3 x}$

Sol:

$\sqrt{\left(8^{0}+\frac{2}{3}\right)}=(0.6)^{2-3 x}$ 

$\left(8^{\circ}=1\right)$

$\sqrt{1+\frac{2}{3}}=\left(\frac{6}{10}\right)^{2-3 x}$

$\sqrt{\left(\frac{5}{3}\right)}=\left(\frac{3}{5}\right)^{2-3 x}$

$\left(\frac{5}{3}\right)^{1 / 2}=\left(\frac{5}{3}\right)^{-(2-3 x)}$

$\left(\frac{5}{3}\right)^{1 / 2}=\left(\frac{5}{3}\right)^{-2+3 x}$

$-2+3 x=\frac{1}{2} \quad$ [comparing in power]

$=3 x=\frac{1}{2}+2=\frac{5}{2}$

$=x=\frac{5}{2 \times 3}=\frac{5}{6}$

$\therefore x=\frac{5}{6}$

Question 11

$3^{2 x+4}+1=2 \cdot 3^{x+2}$

Sol:

$3^{2 x+4}+1=2 \cdot 3^{x+2}$

$3^{2 x} \cdot 3^{4}+1=2 \cdot 3^{x} \cdot 3^{2}$

$81 \cdot 3^{2 x}+1=18 \cdot 3^{x}$

$81 \cdot 32 x-18 \cdot 3^{x}+1=0$

$3 x=a$,

$3^{2 x}=a^{2}$

$\begin{aligned} \therefore & 81 a^{2}-18 a+1=0 \\ &(9 a)^{2}-2 \times 9 a+1=0 \\ &(9 a-1)^{2}=0 \\ &: 9 a-1=0 \\ \Rightarrow & 9 a=1=0 \\ \Rightarrow & 9 a=1 \\ \Rightarrow & 9 \cdot 3^{x}=1 \end{aligned}$

$3^{x}=\frac{1}{9}=3^{-2}$  If comparing both sides power

x=-2 ans.

Question 12

$5^{2 x+1}=6 \cdot 5^{x}-1 $

Sol:

$\begin{aligned} & 5^{2 x+1}=6 \cdot 5^{x}-1 \\=& 5^{2 x} \cdot 5^{2}-6 \cdot 5^{x}+1=0 \\ & 5 \cdot 5^{2 x}-6 \cdot 5^{x}+1=0 \end{aligned}$

$\begin{aligned} 5^{x} &=a \\ \text { and } 5^{2 x} &=a^{2} \end{aligned}$

$\begin{aligned} \therefore \quad 5 a^{2}-6 a+1 &=0 \\ 5 a^{2}-5 a-a+1 &=0 \\ 5 a(a-1)-1(a-1) &=0 \\(a-1)(5 a-1) &=0 \end{aligned}$

$\Rightarrow$

=$\begin{aligned}a-1 &=0 \\a &=2\end{aligned}$

$5 a-1=0$

$5 a=1$

$a=\frac{1}{5}$

case (i)

$\begin{aligned}&a=1,5^{x}=1=5^{\circ} \\&\therefore x=0\end{aligned}$

case (ii) $a=\frac{1}{3}, \quad 5^{x}=\frac{1}{5}=5^{-L}$ $\therefore x=-1$

$\therefore x=-1$

$\begin{aligned} \text { Hence } x &=0 \\ x &=-1 \end{aligned}$

Question 13

$2^{2 x}-2^{x+3}=-2^{4}$

Sol:

$\begin{aligned} & 2^{2 x}-2^{x+3}=-2^{4} \\ & 2^{2 x}-2^{x} \cdot 2^{3}+2^{4}=0 \\ & 2^{2 x}-8 \cdot 2 x+16=0 \\ \because & 2^{x}=a, \quad 2^{2 x}=a^{2} \\ & a^{2}-8 a+16=0 \\ \Rightarrow &(a)^{2}-2 \times a \times 4+(4)^{2}=0 \end{aligned}$

$=(a-4)^{2}=0$

$a-4=0$

$a=2 x=4$

$2 x=2^{2}$(Comparing)

x=2 Ans

Question 14

$9^{x}=3^{y-2}, 81^{y}=3^{2} \times(27)^{x}$

Sol:

$\begin{aligned}&9^{x}=3^{y-2}, 81^{y}=3^{2} \times(27)^{x} \\&{\left[(3)^{2}\right]^{x}=(3)^{y-2}} \\&\Rightarrow 3^{2 x}=3 y^{-2}\end{aligned}$

comparing both sides power.

$2 x=y-2$

$y=2 x+2$............(equal -1)

$\therefore 81^{y}=3^{2 \times(27)^{x}}$

$\Rightarrow\left(3^{4}\right)^{y}=3^{2}\left(3^{3}\right)^{x} \quad(3 \times 3 \times 3 \times 3=81)$

$\begin{aligned}\left(3^{4}\right)^{y} &=3^{2} \times 3^{3 x} \\-3^{4 y} &=3^{3 x+2} \\ \therefore 4 y &=3 x+2 \end{aligned}$........(equal -2)

from equation -1

$a(2 x+2)=3 x-2$

$8 x+8=3 x+2$

$3 x-3 x=2-8$

$5 x=-6$

$x=\frac{-6}{5}$

$\therefore y=2 x+2=2 \times\left(-\frac{6}{5}\right)+2$

$=\frac{-12}{5}+2=\frac{-12+10}{5}=\frac{-2}{5}$

$\therefore x=\frac{-6}{5}$

$\therefore y=\frac{-2}{5}$

Hence proved

Question 15

$2^{1-\frac{x}{2}}=4^{y}$ , $7^{1+x} \times(49)^{-2 y}=1$

Sol:

$2^{1-\frac{x}{2}}=4^{y} \Rightarrow 2^{1-\frac{x}{2}}=2^{2 y}$

$1-\frac{x}{2}=2 y$

$2-x=4 y$

$4 y+x=2$

$x=2-4 y$...........(eq 2)

as $7^{1+x} \times(49)^{-2 y}=1$

$7^{1+x} \cdot\left((7)^{2}\right)^{-2 y}=1$.........(1=7)

$\begin{aligned} 7^{1+x} \cdot 7^{-4 y} &=7^{\circ} \\ 7^{1+x-4 y} &=7^{\circ} \\ \therefore \quad 1+x-4 y &=0 \\ 1+(2-4 y) &-4 y=0 \end{aligned}$

$\therefore \quad 1+x-4 y=0$

$\Rightarrow 1+(2-4 y)-4 y=0$

From equation 1

$x=2-4 y$

$1+2-4 y-4 y=0$

$3-8 y=0$

$8 y=3 \Rightarrow y=3 / 8 .$

$x=2-4 y=2-4 \times \frac{3}{8}$

$=2-3 / 2$

$=\frac{4-3}{2}=\frac{1}{2}$

Hence 

$x=\frac{1}{2}$

$y=3 / 8$

Question 16

$2^{x}=16 \times 2^{y},(27)^{x}=9 \times 3^{2 y}$

Sol:

$\begin{aligned} 2^{x} &=16 \times 2^{y},(27)^{x}=9 \times 3^{2 y} \\ 2^{x} &=16 \times 2^{y} \\ 2^{x} &=2^{4} \times 2 y \\ 2^{x} &=2^{4}+y \end{aligned}$

Comparing both sides power

$\therefore x=4+y$............(eq-1)

 $\begin{aligned}(27)^{x} &=9 \times 3^{2 y} \\\left(3^{3}\right)^{x} &=3^{2} \times 3^{2} y \\ \therefore 3^{3 x} &=3^{2 y+2} \end{aligned}$

$3 x=2 y+1$ (3 Common )

$3(4+y)=2 y+2$ From Eq-1

$12+3 y=2 y+2$

$3 y-2 y=2-12$

$y=-10$

$\begin{aligned} \therefore x=4+y &=4-10 \\ &=-6 \end{aligned}$

$\left[\begin{array}{l}x=-6 \\ y=-10\end{array}\right]$ Hence proved