RD Sharma Solutions for Class 9 Maths Chapter 2 Exponents of Real Numbers – Free PDF Download

# RD Sharma Solutions for Class 9 Maths Chapter 2 Exponents of Real Numbers – Free PDF Download

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## RD Sharma Solutions for Class 9 Maths Chapter 2 Exponents of Real Numbers – Free PDF Download

Question 1: Simplify the following

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

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

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

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

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

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

Solution. Using laws: $\left(a^{m}\right)=a^{m n}, a^{0}=1, a^{m}=1 / a$ and $\left.a^{m} \times a^{n}=a^{m+n}\right]$ (i) $3\left(a^{4} b^{3}\right)^{10} \times 5\left(a^{2} b^{2}\right)^{3}$

On simplifying the given equation, we get;

$=3\left(a^{40} b^{30}\right) \times 5\left(a^{\circ} b^{c}\right)$

$=15\left(a^{40} b^{36}\right)$

[using laws: $\left(a^{m}\right)^{n}=a^{m n}$ and $\left.a^{m} \times a^{n}=a^{m+n}\right]$

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

On simplifying the given equation, we get;

$=\left(2^{3} \mathrm{X}^{-2 \times 3} \mathrm{y}^{3 \times 3}\right)$

$=8 x^{-0} y^{9}$

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

$=\frac{\left(24 \times 10^{7} \times 10^{-5}\right)}{8 \times 10^{4}}$

$=\frac{\left(24 \times 10^{7-5}\right)}{8 \times 10^{4}}$

$=\frac{\left(24 \times 10^{2}\right)}{8 \times 10^{4}}$

$=\frac{\left(3 \times 10^{2}\right)}{10^{4}}$

$=\frac{3}{100}$

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

$=\frac{4 \times(-5)}{10}$

$\times a^{1+1-2} b^{2+3-2}$

$=-2 \times a^{0} b^{3}$

$=-2 b^{3}$

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

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

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

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

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

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

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

$=a^{16 n-50}$

Question 2: If $a=3$ and $b=-2$, find the values of:

(i) $\mathbf{a}^{a}+b^{b}$

(ii) $\mathbf{a}^{\mathrm{b}}+\mathbf{b}^{\mathrm{a}}$

(iii) $(\mathrm{a}+\mathrm{b})^{\mathrm{ab}}$

Solution. (i) $a^{a}+b^{b}$

Now putting the values of ‘ $a$ ‘ and ‘ $b$ ‘, we get;

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

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

$=27+1 / 4$

$=109 / 4$ (ii) $a^{b}+b^{a}$

Now putting the values of ‘a’ and ‘ $b$ ‘, we get;

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

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

$=1 / 9-8$

$=-71 / 9$

(iii) $(\mathrm{a}+\mathrm{b})^{\mathrm{ab}}$

Now putting the values of ‘a’ and ‘ $b$ ‘, we get;

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

$=(3-2))^{-6}$

$=1^{-6}$

$=1$

Question 3: Prove that

Solution. (i) $L . H . S .=$

$\frac{x^{a^{3}+a^{2} b+a b^{2}}}{x^{a^{2} b+a b^{2}+b^{3}}}$

$\times \frac{x^{b^{3}+b^{2} c+b c^{2}}}{x^{b^{2} c+b c^{2}+c^{3}}}$

$\times \frac{x^{c^{3}+c^{2} a+c a^{2}}}{x^{c^{2} a+c a^{2}+a^{3}}}$

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

$\times x^{b^{3}+b^{2} c+b c^{2}-\left(c^{3}+b^{2} c+b c^{2}\right)}$

$\times x^{c^{3}+c^{2} a+c a^{2}-\left(a^{3}+c^{2} a+c a^{2}\right)}$

$=x^{a^{3}-b^{3}} \times x^{b^{3}-c^{3}} \times x^{c^{3}-a^{3}}$

$=x^{a^{3}-b^{3}+b^{3}-c^{3}+c^{3}-a^{3}}$

$=x^{0}$

$=1$

= R.H.S.

(ii) We have to prove here;

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

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

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

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

L.H.S. $=$

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

$\times x^{(b+c)\left(b^{2}-b c+c^{2}\right)}$

$\times x^{(c+a)\left(c^{2}-c a+a^{2}\right)}$

$=x^{a^{3}+b^{3}} \times x^{b^{3}+c^{3}} \times x^{c^{3}+a^{3}}$ $=x^{a^{3}+b^{3}+b^{3}+c^{3}+c^{3}+a^{3}}$

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

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

(iii) L.H.S. =

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

$=\left(\frac{x^{a c}}{x^{b c}}\right) \times\left(\frac{x^{b a}}{x^{c a}}\right) \times\left(\frac{x^{b c}}{x^{a b}}\right)$

$=x^{a c-b c} \times x^{b a-c a} \times x^{b c-a b}$

$=x^{a c-b c+b a-c a+b c-a b}$

$=x^{0}$

$=1$

Question 4: Prove that

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

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

Solution: (i) L.H.S

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

$=\frac{x^{b}}{x^{b}+x^{a}}+\frac{x^{a}}{x^{a}+x^{b}}$

$=\frac{x^{b}+x^{a}}{x^{a}+x^{b}}$

$=1$

= R.H.S.

(ii) L.H.S

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

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

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

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

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

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

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

$=1$

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

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

Solution: (i) L.H.S.

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

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

$=a b c$

= R.H.S.

(ii) L.H.S.

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

$=\frac{1}{\left(\frac{1}{a}+\frac{1}{b}\right)}$

$=\frac{1}{\left(\frac{a+b}{a b}\right)}$

$=\frac{a b}{a+b}$

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

Question 6: If $a b c=1$, show that

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

Solution: $=\frac{1}{1+a+\frac{1}{b}}$

$+\frac{1}{1+b+\frac{1}{c}}$

$+\frac{1}{1+c+\frac{1}{a}}$

$=\frac{b}{b+a b+1}$

$+\frac{c}{c+b c+1}$

$+\frac{a}{a+a c+1}$$\ldots(1)$

Given, $a b c=1$

So, $c=1 / a b$

By putting the value $c$ in equation (1)

$=\frac{b}{b+a b+1}$

$+\frac{\frac{1}{a b}}{\frac{1}{a b}+b\left(\frac{1}{a b}\right)+1}$

$+\frac{a}{a+a\left(\frac{1}{a b}\right)+1}$c

$=\frac{b}{b+a b+1}$

$+\frac{\frac{1}{a b} \times a b}{1+b+a b}$

$+\frac{a b}{1+a b+b}$

$=\frac{b}{b+a b+1}$

$+\frac{1}{1+b+a b}$

$+\frac{a b}{1+a b+b}$

$=\frac{1+a b+b}{b+a b+1}$

$=1$

Exercise $2.2$

Question 1: Assuming that $\mathrm{x}, \mathrm{y}, \mathrm{z}$ are positive real numbers, simplify each of the following:

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

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

(iii) $\left(x^{-\frac{2}{3}} y^{-\frac{1}{2}}\right)^{2}$

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

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

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

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

Solution: (i) $\left(\sqrt{\left(x^{-3}\right)}\right)^{5}=\left(\sqrt{\frac{1}{x^{3}}}\right)^{5}$

$\left(\frac{1}{x^{\frac{3}{2}}}\right)^{5}=\frac{1}{x^{\frac{15}{2}}}$

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

(iii) $\left(x^{-\frac{2}{3}} y^{-\frac{1}{2}}\right)^{2}$

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

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

$=\frac{1}{x^{\frac{4}{3}} y}$

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

$=\left(x^{\frac{1}{2}}\right)^{-\frac{2}{3}}\left(y^{2}\right) \div \sqrt{x y^{-\frac{1}{2}}}$

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

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

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

$=\left(x^{\frac{-2-3}{6}}\right)\left(y^{\frac{8+1}{4}}\right)$

$=\left(x^{-\frac{5}{6}}\right)\left(y^{\frac{9}{4}}\right)$

$=\frac{y^{\frac{9}{4}}}{x^{\frac{5}{6}}}$

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

$=\left(243 x^{10} y^{5} z^{10}\right)^{\frac{1}{5}}$

$=(243)^{\frac{1}{5}} x^{\frac{10}{5}} y^{\frac{5}{5}} z^{\frac{10}{5}}$

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

$=3 x^{2} y z^{2}$

$(\mathrm{Vi})\left(\frac{x^{-4}}{y^{-10}}\right)^{\frac{5}{4}}$

$=\left(\frac{y^{10}}{x^{4}}\right)^{\frac{5}{4}}$

$=\left(\frac{y^{10 \times \frac{5}{4}}}{x^{4 \times \frac{5}{4}}}\right)$

$=\left(\frac{y^{\frac{25}{2}}}{x^{5}}\right)$

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

$=\left(\sqrt{\frac{2}{3}}\right)^{5}\left(\frac{6}{7}\right)^{\frac{4}{2}}$

$=\left(\frac{2}{3}\right)^{\frac{5}{2}}\left(\frac{6}{7}\right)^{\frac{4}{2}}$

$=\left(\frac{2^{5}}{3^{5}}\right)^{\frac{1}{2}}\left(\frac{6^{4}}{7^{4}}\right)^{\frac{1}{2}}$

$=\left(\frac{2^{5}}{3^{5}} \times \frac{6^{4}}{7^{4}}\right)^{\frac{1}{2}}$

$=\left(\frac{2 \times 2 \times 2 \times 2 \times 2}{3 \times 3 \times 3 \times 3 \times 3} \times \frac{6 \times 6 \times 6 \times 6}{7 \times 7 \times 7 \times 7}\right)$

$=\left(\frac{512}{7203}\right)^{\frac{1}{2}}$

Question 2: Simplify

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

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

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

Solution: (i) $\left(16^{-\frac{1}{5}}\right)^{\frac{5}{2}}$

$=(16)^{-\frac{1}{5} \times \frac{5}{2}}=(16)^{-\frac{1}{2}}$

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

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

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

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

$=\left(2^{-15}\right)^{\frac{1}{5}}$

$=2^{-3}$

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

$=\frac{1}{8}$

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

$=\left[(343)^{-2}\right]^{\frac{1}{3}}$

$=(343)^{-2 \times \frac{1}{3}}$

$=\left(7^{3}\right)^{-\frac{2}{3}}$

$=\left(7^{-2}\right)$

$=\left(\frac{1}{7^{2}}\right)$

$=\left(\frac{1}{49}\right)$

(iv) $(0.001)^{\frac{1}{3}}$

$=\left(\frac{1}{10^{3}}\right)^{\frac{1}{3}}=\frac{1}{10^{3 \times \frac{1}{3}}}$

$=\frac{1}{10}=0.1$

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

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

$=\frac{5^{3} \times 3^{3}}{2^{5} \times 2^{4}}$

$=\frac{125 \times 27}{32 \times 16}$

$=\frac{3375}{512}$

(vi) $\left(\frac{\sqrt{2}}{5}\right)^{8} \div\left(\frac{\sqrt{2}}{5}\right)^{13}=\left(\frac{\sqrt{2}}{5}\right)^{8-13}$

$=\left(\frac{\sqrt{2}}{5}\right)^{-5}=\frac{5^{5}}{2^{\frac{5}{2}}}=\frac{3125}{4 \sqrt{2}}$

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

$=\frac{5^{-1 \times \frac{7}{2}} \times 7^{2 \times \frac{7}{2}}}{5^{2 \times \frac{7}{2}} \times 7^{-4 \times \frac{7}{2}}} \times \frac{5^{-2 \times\left(\frac{-5}{2}\right)} \times 7^{3 \times\left(\frac{-5}{2}\right)}}{5^{3 \times\left(\frac{-5}{2}\right)} \times 7^{-5 \times\left(\frac{-5}{2}\right)}}$

$=\frac{5^{\frac{-7}{2}} \times 7^{7}}{5^{7} \times 7^{-14}}=\frac{5^{5} \times 7^{\frac{-15}{2}}}{5^{\frac{-15}{2}} \times 7^{\frac{25}{2}}}$

$=5^{\frac{-7}{2}+5-7+\frac{15}{2}} \times 7^{7-\frac{15}{2}+14-\frac{25}{2}}$

$=5^{\frac{25}{2}}-\frac{21}{2} \times 7^{21-\frac{40}{2}}=5^{\frac{4}{2}} \times 7^{\frac{2}{2}}$

$=5^{2} \times 7^{1}=25 \times 7=175$

Question 3: Prove that

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

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

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

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

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

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

(vii) $\left(\frac{64}{125}\right)^{-\frac{2}{3}}+\frac{1}{\left(\frac{256}{625}\right)^{\frac{1}{4}}}+\left(\frac{\sqrt{25}}{\sqrt[3]{64}}\right)^{0}=\frac{61}{16}$

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

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

Solution: (i) L.H.S.

$\left(\sqrt{3 \times 5^{-3}} \div \sqrt[3]{3^{-1}} \sqrt{5}\right) \times \sqrt[6]{3 \times 5^{6}}$

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

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

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

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

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

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

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

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

$=\left((3)^{\frac{3+2}{6}} \times(5)^{-\frac{4}{2}}\right)$

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

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

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

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

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

$=\left((3)^{1} \times(5)^{-1}\right)$

$=\left((3) \times(5)^{-1}\right)$

$=\left((3) \times\left(\frac{1}{5}\right)\right)$

$=\left(\frac{3}{5}\right)$

= R.H.S.

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

$=\left(3^{2}\right)^{\frac{3}{2}}-3-\left(\frac{1}{9^{2}}\right)^{-\frac{1}{2}}$

$=3^{2 \times \frac{3}{2}}-3-\left(9^{-2}\right)^{-\frac{1}{2}}$

$=3^{3}-3-(9)^{-2 \times-\frac{1}{2}}$

$=27-3-9$

$=15$

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

$=\left(\frac{1}{2^{2}}\right)^{-2}-3 \times 8^{\frac{2}{3}} \times 1+\left(\frac{3^{2}}{4^{2}}\right)^{-\frac{1}{2}}$

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

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

$=\frac{12+4}{3}$

$=\frac{16}{3}$

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

$=\frac{2^{\frac{1}{2}} \times 3^{\frac{1}{3}} \times\left(2^{2}\right)^{\frac{1}{4}}\left(2^{2}\right)^{-\frac{3}{5}} \times(2 \times 3)}{(2 \times 5)^{-\frac{1}{5}} \times 5^{\frac{3}{5}} \times 3^{\frac{4}{3}} \times 5^{-\frac{7}{5}}}$

$=\frac{2^{\frac{1}{2}} \times 2^{\frac{1}{2}} \times\left(2^{2}\right)^{-\frac{6}{5}} \times 2^{1} \times 3^{\frac{1}{3}} \times 3}{2^{-\frac{1}{5}} \times 5^{-\frac{1}{5}} \times 5^{\frac{3}{5}} \times 3^{\frac{4}{3}} \times 5^{-\frac{7}{5}}}$

$=\frac{2^{\frac{1}{5}} \times 2^{\frac{1}{2}} \times 2^{\frac{1}{2}} \times 2^{-\frac{6}{5}} \times 2 \times 3^{\frac{1}{3}} \times 3 \times 3^{-\frac{4}{3}}}{5^{-\frac{1}{5}} \times 5^{\frac{3}{5}} \times 5^{-\frac{7}{5}}}$

$=\frac{(2)^{\frac{1}{2}+\frac{1}{2}-\frac{6}{5}+1+\frac{1}{5}} \times(3)^{\frac{1}{3}+1-\frac{4}{3}}}{5^{-\frac{1}{5}} \times 5^{\frac{3}{5}} \times 5^{-\frac{7}{5}}}$

$=\frac{(2)^{\frac{1}{5}+2-\frac{6}{5}} \times(3)^{1-1}}{5^{-1}}$

$=\frac{(2)^{1} \times(3)^{0}}{5^{-1}}$

$=2 \times 1 \times 5$

$=10$

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

$=\frac{1}{2}+\frac{1}{(0.01)^{\frac{1}{2}}}-\left(3^{3}\right)^{\frac{2}{3}}$

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

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

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

$=\frac{3}{2}$

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

$=\frac{2^{n}+2^{n} \times 2^{-1}}{2^{n} \times 2^{1}-2^{n}}$

$=\frac{2^{n}\left[1+2^{-1}\right]}{2^{n}[2-1]}$

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

$=\frac{3}{2}$

(vii) $\left(\frac{64}{125}\right)^{-\frac{2}{3}}+\frac{1}{\left(\frac{256}{625}\right)^{\frac{1}{4}}}+\left(\frac{\sqrt{25}}{\sqrt[3]{64}}\right)^{0}$

$=\left(\frac{125}{64}\right)^{\frac{2}{3}}+\frac{1}{\left(\frac{4^{4}}{5^{4}}\right)^{\frac{1}{4}}}+1$

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

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

$=\frac{25}{16}+\frac{9}{4}$

$=\frac{25}{16}+\frac{36}{16}$

$=\frac{61}{16}$

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

$=\frac{3^{-3} \times 36 \times \sqrt{7 \times 7 \times 2}}{5^{2} \times\left(\frac{1}{25}\right)^{\frac{1}{3}} \times(15)^{-\frac{4}{3}} \times 3^{\frac{1}{3}}}$

$=\frac{3^{-3} \times 2^{2} \times 3^{2} \times 2^{\frac{1}{2}} \times\left(7^{2}\right)^{\frac{1}{2}}}{5^{2} \times\left(5^{2}\right)^{\frac{-1}{3}} \times 3^{\frac{-4}{3}} \times 5^{\frac{-4}{3}} \times 3^{\frac{1}{3}}}$

$=2^{2} \cdot 2^{\frac{1}{2}} \cdot 3^{-3+2+\frac{4}{3}-\frac{1}{3}} \cdot 5^{\frac{2}{3}-2+\frac{4}{3}} \cdot 7^{1}$

$=4 \sqrt{2} \times 3^{0 \times} 5^{0} \times 7^{1}$

$=4 \sqrt{2} \times 1 \times 1 \times 7$

$=28 \sqrt{2}$

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

$=\frac{1-\frac{1}{0.1}}{\frac{8}{3} \times\left(\frac{3}{2}\right)^{3}-3}$

$=\frac{1-10}{\frac{8}{3} \times \frac{3^{3}}{2^{3}}-3}$

$=\frac{-9}{3^{2}-3}$

$=-3 / 2$

Question $4 .$ Show that:

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

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

$=\frac{x^{b}}{x^{b}+x^{a}}+\frac{x^{a}}{x^{a}+x^{b}}$

$=\frac{x^{b}+x^{a}}{x^{a}+x^{b}}$

$=1$

$(i i)\left[\left(\frac{x^{a(a-b)}}{x^{a(a+b)}}\right) \div\left(\frac{x^{b(b-a)}}{x^{b(b+a)}}\right)\right]^{a+b}$

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

$=\left[x^{\left(a^{2}-a b\right)-\left(a^{2}-a b\right)}\right.$

$\left.\div x^{\left(b^{2}-a b\right)-\left(b^{2}-a b\right)}\right]^{a+b}$

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

$=\left[x^{0}\right]^{a+b}=[1]^{a+b}=1$

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

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

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

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

$=x^{0}=1$

$(i v)\left(\frac{x^{a^{2}+b^{2}}}{x^{a b}}\right)^{a+b}$

$\left(\frac{x^{b^{2}+c^{2}}}{x^{b c}}\right)^{b+c}$

$\left(\frac{x^{c^{2}+a^{2}}}{x^{a c}}\right)^{a+c}$

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

$\left(x^{b^{2}+c^{2}-b c}\right)^{b+c}$

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

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

$\left(x^{b+c\left(b^{2}+c^{2}-b c\right)}\right)$

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

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

$\left(x^{b^{3}+b c^{2}-b^{2} c+c b^{2}+c^{3}-b c^{2}}\right)$

$\left(x^{a c^{2}+a^{3}-a^{2} c+c^{3}+a^{2} c-a c^{2}}\right)$

$=\left(x^{a^{3}+b^{3}}\right)\left(x^{b^{3}+c^{3}}\right)\left(x^{a^{3}+c^{3}}\right)$

$=\left(x^{a^{3}+b^{3}+b^{3}+c^{3}+a^{3}+c^{3}}\right)$

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

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

$\left(x^{a-b}\right)^{a+b}\left(x^{b-c}\right)^{b+c}\left(x^{c-a}\right)^{c+a}$

$=x^{a^{2}-b^{2}} x^{b^{2}-c^{2}} x^{c^{2}-a^{2}}$

$=x^{a^{2}-b^{2}+b^{2}-c^{2}+c^{2}-a^{2}}$

$=x^{0}$

$=1$

$(v i)$ $\left[\left(x^{a-a^{-1}}\right)^{\frac{1}{a-1}}\right]^{\frac{a}{a+1}}$

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

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

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

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

$=x^{1}=x$

$(v i i)\left[\frac{a^{x+1}}{a^{y+1}}\right]^{x+y}$

$\left[\frac{a^{y+2}}{a^{z+2}}\right]^{y+z}$

$\left[\frac{a^{z+3}}{a^{x+3}}\right]^{z+x}$

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

$\left[a^{(y+2)-(z+2)}\right]^{y+z}$

$\left[a^{(z+3)-(x+3)}\right]^{z+x}$

$=\left[a^{x-y}\right]^{x+y}$

$\left[a^{y-z}\right]^{y+z}$

$\left[a^{\tilde{z}-x}\right]^{\tilde{z}+x}$

$=\left[a^{x^{2}-y^{2}}\right]$

$\left[a^{y^{2}-z^{2}}\right]$

$\left[a^{z^{2}-x^{2}}\right]$

$=a^{x^{2}-y^{2}+y^{2}-z^{2}+z^{2}-x^{2}}$

$=a^{0}=1$

$(v i i i)\left(\frac{3^{a}}{3^{b}}\right)^{a+b}$

$\left(\frac{3^{b}}{3^{c}}\right)^{b+c}$

$\left(\frac{3^{c}}{3^{a}}\right)^{c+a}$

$=\left(3^{a-b}\right)^{a+b}$

$\left(3^{b-c}\right)^{b+c}$

$\left(3^{c-a}\right)^{c+a}$

$=3^{a^{2}-b^{2}} \times 3^{b^{2}-c^{2}} \times 3^{c^{2}-a^{2}}$

$=3^{a^{2}-b^{2}+b^{2}-c^{2}+c^{2}-a^{2}}$

$=3^{0}=1$

Exercise-VSAQs

Question 1: Write (625) ${ }^{-14}$ in decimal form.

Solution: $(625)^{-1 / 4}=\left(5^{4}\right)^{-14}=5^{-1}=1 / 5=0.2$

Question 2: State the product law of exponents:

Solution: To multiply two parts having same base, add the exponents.

Mathematically: $\mathrm{x}^{\mathrm{m}} \times \mathrm{x}^{\mathrm{n}}=\mathrm{x}^{\mathrm{m}+n}$

Question 3: State the quotient law of exponents.

Solution: To divide two exponents with the same base, subtract the powers.

Mathematically: $x^{m} \div x^{n}=x^{m-n}$

Question 4: State the power law of exponents.

Solution: Power law of exponents :

$\left(\mathrm{X}^{\mathrm{m}}\right)^{\mathrm{n}}=\mathrm{X}^{\mathrm{m} \times \mathrm{n}}=\mathrm{X}^{\mathrm{mn}}$

Question 5: For any positive real number $\mathrm{x}$, find the value of

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

Solution: $\left(\frac{x^{a}}{x^{b}}\right)^{a+b}$

$\left(\frac{x^{b}}{x^{c}}\right)^{b+c}$

$\left(\frac{x^{c}}{x^{a}}\right)^{c+a}$

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

$\times\left(x^{c-a}\right)^{c+a}$

$=x^{a^{2}-b^{2}} \times x^{b^{2}-c^{2}}$

$\times x^{c^{2}-a^{2}}$

$=x^{a^{2}-b^{2}+b^{2}-c^{2}+c^{2}-a^{2}}$

$=1$

Question $6:$ Write the value of $\left\{5\left(8^{1 / 3}+27^{1 / 3}\right)^{3}\right\}^{1 / 4}$

Solution: $\left\{5\left(8^{1 / 3}+27^{1 / 3}\right)^{3}\right\}^{1 / 4}$

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

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

$=\left(5^{4}\right)^{1 / 4}$

$=5$