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#### Question 1:

Simplify the following:

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

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

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

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

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

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

(i)

(ii)
${\left(2{x}^{-2}{y}^{3}\right)}^{3}\phantom{\rule{0ex}{0ex}}={2}^{3}×{\left({x}^{-2}\right)}^{3}×{\left({y}^{3}\right)}^{3}\phantom{\rule{0ex}{0ex}}=8×{x}^{-6}×{y}^{9}\phantom{\rule{0ex}{0ex}}=8{x}^{-6}{y}^{9}$

(iii)
$\frac{\left(4×{10}^{7}\right)\left(6×{10}^{-5}\right)}{8×{10}^{4}}\phantom{\rule{0ex}{0ex}}=\frac{4×{10}^{7}×6×{10}^{-5}}{8×{10}^{4}}\phantom{\rule{0ex}{0ex}}=\frac{24×{10}^{7+\left(-5\right)}}{8×{10}^{4}}\phantom{\rule{0ex}{0ex}}=\frac{24×{10}^{2}}{8×{10}^{4}}$
$=\frac{24×{10}^{2}×{10}^{-4}}{8}\phantom{\rule{0ex}{0ex}}=3×{10}^{2+\left(-4\right)}\phantom{\rule{0ex}{0ex}}=3×{10}^{-2}\phantom{\rule{0ex}{0ex}}=\frac{3}{100}$

(iv)
$\frac{4a{b}^{2}\left(-5a{b}^{3}\right)}{10{a}^{2}{b}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{4×a×{b}^{2}×\left(-5\right)×a×{b}^{3}}{10{a}^{2}{b}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{-20×{a}^{1}×{a}^{1}×{b}^{2}×{b}^{3}}{10{a}^{2}{b}^{2}}$
$=\frac{-20×{a}^{1+1}×{b}^{2+3}}{10{a}^{2}{b}^{2}}\phantom{\rule{0ex}{0ex}}=-2×{a}^{2}×{b}^{5}×{a}^{-2}×{b}^{-2}\phantom{\rule{0ex}{0ex}}=-2×{a}^{2+\left(-2\right)}×{b}^{5+\left(-2\right)}\phantom{\rule{0ex}{0ex}}=-2×{a}^{0}×{b}^{3}\phantom{\rule{0ex}{0ex}}=-2{b}^{3}$

(v)
${\left(\frac{{x}^{2}{y}^{2}}{{a}^{2}{b}^{3}}\right)}^{n}\phantom{\rule{0ex}{0ex}}=\frac{{\left({x}^{2}\right)}^{n}{\left({y}^{2}\right)}^{n}}{{\left({a}^{2}\right)}^{n}{\left({b}^{3}\right)}^{n}}\phantom{\rule{0ex}{0ex}}=\frac{{x}^{2n}{y}^{2n}}{{a}^{2n}{b}^{3n}}$

(vi)
$\frac{{\left({a}^{3n-9}\right)}^{6}}{{a}^{2n-4}}\phantom{\rule{0ex}{0ex}}=\frac{{a}^{6\left(3n-9\right)}}{{a}^{2n-4}}\phantom{\rule{0ex}{0ex}}=\frac{{a}^{\left(18n-54\right)}}{{a}^{2n-4}}$
$={a}^{\left(18n-54\right)}×{a}^{-\left(2n-4\right)}\phantom{\rule{0ex}{0ex}}={a}^{18n-54}×{a}^{-2n+4}\phantom{\rule{0ex}{0ex}}={a}^{18n-54-2n+4}\phantom{\rule{0ex}{0ex}}={a}^{16n-50}$

#### Question 2:

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

(i) ${a}^{a}+{b}^{b}$
Here, $a=3$ and $b=-2$.
Put the values in the expression ${a}^{a}+{b}^{b}$.
${3}^{3}+{\left(-2\right)}^{-2}\phantom{\rule{0ex}{0ex}}=27+\frac{1}{{\left(-2\right)}^{2}}\phantom{\rule{0ex}{0ex}}=27+\frac{1}{4}\phantom{\rule{0ex}{0ex}}=\frac{108+1}{4}\phantom{\rule{0ex}{0ex}}=\frac{109}{4}$

(ii) ${a}^{b}+{b}^{a}$
Here, $a=3$ and $b=-2$.
Put the values in the expression ${a}^{b}+{b}^{a}$.
${3}^{-2}+{\left(-2\right)}^{3}\phantom{\rule{0ex}{0ex}}={\left(\frac{1}{3}\right)}^{2}+\left(-8\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{9}-8\phantom{\rule{0ex}{0ex}}=\frac{1-72}{9}\phantom{\rule{0ex}{0ex}}=-\frac{71}{9}$

(iii) ${\left(a+b\right)}^{ab}$
Here, $a=3$ and $b=-2$.
Put the values in the expression ${\left(a+b\right)}^{ab}$.
${\left[3+\left(-2\right)\right]}^{3\left(-2\right)}\phantom{\rule{0ex}{0ex}}={\left(1\right)}^{-6}\phantom{\rule{0ex}{0ex}}=1$

#### Question 3:

Prove that:

(i) ${\left(\frac{{x}^{a}}{{x}^{b}}\right)}^{{a}^{2}+ab+{b}^{2}}×{\left(\frac{{x}^{b}}{{x}^{c}}\right)}^{{b}^{2}+bc+{c}^{2}}×{\left(\frac{{x}^{c}}{{x}^{a}}\right)}^{{c}^{2}+ca+{a}^{2}}=1$
(ii) ${\left(\frac{{x}^{a}}{{x}^{b}}\right)}^{c}×{\left(\frac{{x}^{b}}{{x}^{c}}\right)}^{a}×{\left(\frac{{x}^{c}}{{x}^{a}}\right)}^{b}=1$

(i) ${\left(\frac{{x}^{a}}{{x}^{b}}\right)}^{{a}^{2}+ab+{b}^{2}}×{\left(\frac{{x}^{b}}{{x}^{c}}\right)}^{{b}^{2}+bc+{c}^{2}}×{\left(\frac{{x}^{c}}{{x}^{a}}\right)}^{{c}^{2}+ca+{a}^{2}}=1$

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

(ii) ${\left(\frac{{x}^{a}}{{x}^{b}}\right)}^{c}×{\left(\frac{{x}^{b}}{{x}^{c}}\right)}^{a}×{\left(\frac{{x}^{c}}{{x}^{a}}\right)}^{b}=1$
Consider the left hand side:
$=\frac{{x}^{ac}}{{x}^{bc}}×\frac{{x}^{ba}}{{x}^{ca}}×\frac{{x}^{cb}}{{x}^{ab}}\phantom{\rule{0ex}{0ex}}=\frac{{x}^{ac}}{{x}^{bc}}×\frac{{x}^{ba}}{{x}^{ca}}×\frac{{x}^{cb}}{{x}^{ab}}\phantom{\rule{0ex}{0ex}}=\frac{{x}^{ac}×{x}^{ba}×{x}^{cb}}{{x}^{bc}×{x}^{ca}×{x}^{ab}}\phantom{\rule{0ex}{0ex}}=\frac{{x}^{ac+ba+cb}}{{x}^{bc+ca+ab}}\phantom{\rule{0ex}{0ex}}=1$
Left hand side is equal to right hand side.
Hence proved.

#### Question 4:

Prove that:

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

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

(i) Consider the left hand side:
$\frac{1}{1+{x}^{a-b}}+\frac{1}{1+{x}^{b-a}}\phantom{\rule{0ex}{0ex}}=\frac{1}{1+\frac{{x}^{a}}{{x}^{b}}}+\frac{1}{1+\frac{{x}^{b}}{{x}^{a}}}\phantom{\rule{0ex}{0ex}}=\frac{1}{\frac{{x}^{b}+{x}^{a}}{{x}^{b}}}+\frac{1}{\frac{{x}^{a}+{x}^{b}}{{x}^{a}}}\phantom{\rule{0ex}{0ex}}=\frac{{x}^{b}}{{x}^{b}+{x}^{a}}+\frac{{x}^{a}}{{x}^{a}+{x}^{b}}\phantom{\rule{0ex}{0ex}}=\frac{{x}^{b}+{x}^{a}}{{x}^{b}+{x}^{a}}\phantom{\rule{0ex}{0ex}}=1$
Therefore left hand side is equal to the right hand side. Hence proved.

(ii)
Consider the left hand side:
$\frac{1}{1+{x}^{b-a}+{x}^{c-a}}+\frac{1}{1+{x}^{a-b}+{x}^{c-b}}+\frac{1}{1+{x}^{b-c}+{x}^{a-c}}\phantom{\rule{0ex}{0ex}}=\frac{1}{1+\frac{{x}^{b}}{{x}^{a}}+\frac{{x}^{c}}{{x}^{a}}}+\frac{1}{1+\frac{{x}^{a}}{{x}^{b}}+\frac{{x}^{c}}{{x}^{b}}}+\frac{1}{1+\frac{{x}^{b}}{{x}^{c}}+\frac{{x}^{a}}{{x}^{c}}}\phantom{\rule{0ex}{0ex}}=\frac{1}{\frac{{x}^{a}+{x}^{b}+{x}^{c}}{{x}^{a}}}+\frac{1}{\frac{{x}^{b}+{x}^{a}+{x}^{c}}{{x}^{b}}}+\frac{1}{\frac{{x}^{c}+{x}^{b}+{x}^{c}}{{x}^{c}}}\phantom{\rule{0ex}{0ex}}=\frac{{x}^{a}}{{x}^{a}+{x}^{b}+{x}^{c}}+\frac{{x}^{b}}{{x}^{b}+{x}^{a}+{x}^{c}}+\frac{{x}^{c}}{{x}^{c}+{x}^{b}+{x}^{c}}\phantom{\rule{0ex}{0ex}}=\frac{{x}^{a}+{x}^{b}+{x}^{c}}{{x}^{a}+{x}^{b}+{x}^{c}}\phantom{\rule{0ex}{0ex}}=1$
Therefore left hand side is equal to the right hand side. Hence proved.

#### Question 5:

Prove that:

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

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

(i) Consider the left hand side:
$\frac{a+b+c}{{a}^{-1}{b}^{-1}+{b}^{-1}{c}^{-1}+{c}^{-1}{a}^{-1}}\phantom{\rule{0ex}{0ex}}=\frac{a+b+c}{\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}}\phantom{\rule{0ex}{0ex}}=\frac{a+b+c}{\frac{c+a+b}{abc}}\phantom{\rule{0ex}{0ex}}=\left(a+b+c\right)×\left(\frac{abc}{a+b+c}\right)\phantom{\rule{0ex}{0ex}}=abc$
Therefore left hand side is equal to the right hand side. Hence proved.

(ii)
Consider the left hand side:
${\left({a}^{-1}+{b}^{-1}\right)}^{-1}\phantom{\rule{0ex}{0ex}}=\frac{1}{{a}^{-1}+{b}^{-1}}\phantom{\rule{0ex}{0ex}}=\frac{1}{\frac{1}{a}+\frac{1}{b}}\phantom{\rule{0ex}{0ex}}=\frac{1}{\frac{b+a}{ab}}\phantom{\rule{0ex}{0ex}}=\frac{ab}{a+b}$
Therefore left hand side is equal to the right hand side. Hence proved.

#### Question 6:

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

Consider the left hand side:

Hence proved.

#### Question 7:

Simplify the following:

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

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

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

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

(i)
$\frac{{3}^{n}×{9}^{n+1}}{{3}^{n-1}×{9}^{n-1}}\phantom{\rule{0ex}{0ex}}=\frac{{3}^{n}×{\left({3}^{2}\right)}^{\left(n+1\right)}}{{3}^{n-1}×{\left({3}^{2}\right)}^{n-1}}\phantom{\rule{0ex}{0ex}}=\frac{{3}^{n}×{3}^{2n+2}}{{3}^{n-1}×{3}^{2n-2}}$
$=\frac{{3}^{n+2n+2}}{{3}^{n-1+2n-2}}\phantom{\rule{0ex}{0ex}}=\frac{{3}^{3n+2}}{{3}^{3n-3}}\phantom{\rule{0ex}{0ex}}={3}^{3n+2-3n+3}\phantom{\rule{0ex}{0ex}}={3}^{5}\phantom{\rule{0ex}{0ex}}=243$

(ii)
$\frac{5×{25}^{n+1}-25×{5}^{2n}}{5×{5}^{2n+3}-{25}^{n+1}}\phantom{\rule{0ex}{0ex}}=\frac{5×{\left({5}^{2}\right)}^{n+1}-\left({5}^{2}\right)×{5}^{2n}}{5×{5}^{2n+3}-{\left({5}^{2}\right)}^{n+1}}\phantom{\rule{0ex}{0ex}}=\frac{5×\left({5}^{2n+2}\right)-\left({5}^{2}\right)×{5}^{2n}}{5×{5}^{2n+3}-\left({5}^{2n+2}\right)}\phantom{\rule{0ex}{0ex}}=\frac{{5}^{1+2n+2}-{5}^{2+2n}}{{5}^{1+2n+3}-{5}^{2n+2}}$
$=\frac{{5}^{2n+3}-{5}^{2+2n}}{{5}^{2n+4}-{5}^{2n+2}}\phantom{\rule{0ex}{0ex}}=\frac{{5}^{2+2n}\left(5-1\right)}{{5}^{2n+2}\left({5}^{2}-1\right)}\phantom{\rule{0ex}{0ex}}=\frac{4}{24}\phantom{\rule{0ex}{0ex}}=\frac{1}{6}$

(iii)
$\frac{{5}^{n+3}-6×{5}^{n+1}}{9×{5}^{n}-{2}^{2}×{5}^{n}}\phantom{\rule{0ex}{0ex}}=\frac{{5}^{n+1}\left({5}^{2}-6\right)}{{5}^{n}\left(9-{2}^{2}\right)}\phantom{\rule{0ex}{0ex}}=\frac{{5}^{n}×5×\left(25-6\right)}{{5}^{n}\left(9-4\right)}\phantom{\rule{0ex}{0ex}}=\frac{5×19}{5}\phantom{\rule{0ex}{0ex}}=19$

(iv)
$\frac{6{\left(8\right)}^{n+1}+16{\left(2\right)}^{3n-2}}{10{\left(2\right)}^{3n+1}-7{\left(8\right)}^{n}}\phantom{\rule{0ex}{0ex}}=\frac{6{\left({2}^{3}\right)}^{n+1}+16{\left(2\right)}^{3n-2}}{10{\left(2\right)}^{3n+1}-7{\left({2}^{3}\right)}^{n}}\phantom{\rule{0ex}{0ex}}=\frac{6\left({2}^{3n+3}\right)+16{\left(2\right)}^{3n-2}}{10{\left(2\right)}^{3n+1}-7\left({2}^{3n}\right)}$
$=\frac{6×{2}^{3n}\left({2}^{3}\right)+16{\left(2\right)}^{3n}{2}^{-2}}{10{\left(2\right)}^{3n}\left({2}^{1}\right)-7\left({2}^{3n}\right)}\phantom{\rule{0ex}{0ex}}=\frac{{2}^{3n}\left(48+4\right)}{{2}^{3n}\left(20-7\right)}\phantom{\rule{0ex}{0ex}}=\frac{52}{13}\phantom{\rule{0ex}{0ex}}=4$

#### Question 8:

Solve the following equations for x:

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

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

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

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

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

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

(i)
${7}^{2x+3}=1\phantom{\rule{0ex}{0ex}}⇒{7}^{2x+3}={7}^{0}\phantom{\rule{0ex}{0ex}}⇒2x+3=0\phantom{\rule{0ex}{0ex}}⇒2x=-3\phantom{\rule{0ex}{0ex}}⇒x=-\frac{3}{2}$

(ii)
${2}^{x+1}={4}^{x-3}\phantom{\rule{0ex}{0ex}}⇒{2}^{x+1}={\left({2}^{2}\right)}^{x-3}\phantom{\rule{0ex}{0ex}}⇒{2}^{x+1}=\left({2}^{2x-6}\right)\phantom{\rule{0ex}{0ex}}⇒x+1=2x-6\phantom{\rule{0ex}{0ex}}⇒x=7$

(iii)
${2}^{5x+3}={8}^{x+3}\phantom{\rule{0ex}{0ex}}⇒{2}^{5x+3}={\left({2}^{3}\right)}^{x+3}\phantom{\rule{0ex}{0ex}}⇒{2}^{5x+3}={2}^{3x+9}\phantom{\rule{0ex}{0ex}}⇒5x+3=3x+9\phantom{\rule{0ex}{0ex}}⇒2x=6\phantom{\rule{0ex}{0ex}}⇒x=3\phantom{\rule{0ex}{0ex}}$

(iv)
${4}^{2x}=\frac{1}{32}\phantom{\rule{0ex}{0ex}}⇒{\left({2}^{2}\right)}^{2x}=\frac{1}{{2}^{5}}\phantom{\rule{0ex}{0ex}}⇒{2}^{4x}×{2}^{5}=1\phantom{\rule{0ex}{0ex}}⇒{2}^{4x+5}={2}^{0}\phantom{\rule{0ex}{0ex}}⇒4x+5=0\phantom{\rule{0ex}{0ex}}⇒x=-\frac{5}{4}$

(v)

(vi)
${2}^{3x-7}=256\phantom{\rule{0ex}{0ex}}⇒{2}^{3x-7}={2}^{8}\phantom{\rule{0ex}{0ex}}⇒3x-7=8\phantom{\rule{0ex}{0ex}}⇒3x=15\phantom{\rule{0ex}{0ex}}⇒x=5\phantom{\rule{0ex}{0ex}}$

#### Question 9:

Solve the following equations for x:

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

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

(i)
${2}^{2x}-{2}^{x+3}+{2}^{4}=0\phantom{\rule{0ex}{0ex}}⇒{\left({2}^{x}\right)}^{2}-\left({2}^{x}×{2}^{3}\right)+{\left({2}^{2}\right)}^{2}=0\phantom{\rule{0ex}{0ex}}⇒{\left({2}^{x}\right)}^{2}-2×{2}^{x}×{2}^{2}+{\left({2}^{2}\right)}^{2}=0\phantom{\rule{0ex}{0ex}}⇒{\left({2}^{x}-{2}^{2}\right)}^{2}=0\phantom{\rule{0ex}{0ex}}⇒{2}^{x}-{2}^{2}=0\phantom{\rule{0ex}{0ex}}⇒{2}^{x}={2}^{2}\phantom{\rule{0ex}{0ex}}⇒x=2$

(ii)
${3}^{2x+4}+1=2.{3}^{x+2}\phantom{\rule{0ex}{0ex}}⇒{\left({3}^{x+2}\right)}^{2}-2.{3}^{x+2}+1=0\phantom{\rule{0ex}{0ex}}⇒{\left({3}^{x+2}-1\right)}^{2}=0\phantom{\rule{0ex}{0ex}}⇒{3}^{x+2}-1=0\phantom{\rule{0ex}{0ex}}⇒{3}^{x+2}=1={3}^{0}\phantom{\rule{0ex}{0ex}}⇒x+2=0\phantom{\rule{0ex}{0ex}}⇒x=-2\phantom{\rule{0ex}{0ex}}$

#### 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.

First find out the prime factorisation of 49392.

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

It can be observed that 49392 can be written as ${2}^{4}×{3}^{2}×{7}^{3}$, where 2, 3 and 7 are positive primes.

#### Question 11:

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

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

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

#### Question 12:

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

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

(ii)
When a = 3, b = 2 and c = 1,
${2}^{-a}{3}^{b}{7}^{c}\phantom{\rule{0ex}{0ex}}={2}^{-3}×{3}^{2}×{7}^{1}\phantom{\rule{0ex}{0ex}}=\frac{1}{8}×9×7\phantom{\rule{0ex}{0ex}}=\frac{63}{8}$
Hence, the value of ${2}^{-a}{3}^{b}{7}^{c}$ is $\frac{63}{8}$.

#### Question 13:

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

It is given that .
$\therefore {a}^{q-r}{b}^{r-p}{c}^{p-q}\phantom{\rule{0ex}{0ex}}={\left(x{y}^{p-1}\right)}^{q-r}{\left(x{y}^{q-1}\right)}^{r-p}{\left(x{y}^{r-1}\right)}^{p-q}\phantom{\rule{0ex}{0ex}}={x}^{\left(q-r\right)}{y}^{\left(p-1\right)\left(q-r\right)}{x}^{\left(r-p\right)}{y}^{\left(r-p\right)\left(q-1\right)}{x}^{\left(p-q\right)}{y}^{\left(p-q\right)\left(r-1\right)}\phantom{\rule{0ex}{0ex}}={x}^{\left(q-r\right)}{x}^{\left(r-p\right)}{x}^{\left(p-q\right)}{y}^{\left(p-1\right)\left(q-r\right)}{y}^{\left(r-p\right)\left(q-1\right)}{y}^{\left(p-q\right)\left(r-1\right)}\phantom{\rule{0ex}{0ex}}={x}^{\left(q-r\right)+\left(r-p\right)+\left(p-q\right)}{y}^{\left(p-1\right)\left(q-r\right)+\left(r-p\right)\left(q-1\right)+\left(p-q\right)\left(r-1\right)}\phantom{\rule{0ex}{0ex}}={x}^{q-r+r-p+p-q}{y}^{pq-q-pr+r+rq-r-pq+p+pr-p-qr+q}\phantom{\rule{0ex}{0ex}}={x}^{0}{y}^{0}\phantom{\rule{0ex}{0ex}}=1$

#### Question 1:

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

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

(ii)

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

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

(v)

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

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

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

By using law of rational exponents we have

${\left({x}^{\frac{-2}{3}}{y}^{{}^{\frac{-1}{2}}}\right)}^{2}=\frac{1}{{x}^{\frac{2}{3}+\frac{2}{3}}}×\frac{1}{{y}^{\frac{1}{2}+\frac{1}{2}}}\phantom{\rule{0ex}{0ex}}=\frac{1}{{x}^{\frac{4}{3}}}×\frac{1}{{y}^{\frac{2}{2}}}=\frac{1}{{x}^{\frac{4}{3}}}×\frac{1}{y}\phantom{\rule{0ex}{0ex}}=\frac{1}{{x}^{\frac{4}{3}}y}$

Hence the simplified value of is .

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

${\left(\frac{\sqrt{2}}{\sqrt{3}}\right)}^{5}{\left(\frac{6}{7}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(\frac{\sqrt{2}}{\sqrt{3}}\right)}^{2+2+1}{\left(\frac{6}{7}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(\frac{\sqrt{2}}{\sqrt{3}}\right)}^{2}×{\left(\frac{\sqrt{2}}{\sqrt{3}}\right)}^{2}×{\left(\frac{\sqrt{2}}{\sqrt{3}}\right)}^{1}×{\left(\frac{6}{7}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(\frac{2}{3}\right)×\left(\frac{2}{3}\right)×{\left(\frac{\sqrt{2}}{\sqrt{3}}\right)}^{1}×{\left(\frac{6}{7}\right)}^{2}$
$=\left(\frac{2}{3}\right)×\left(\frac{2}{3}\right)×{\left(\frac{\sqrt{2}}{\sqrt{3}}\right)}^{1}×{\left(\frac{6}{7}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{16\sqrt{2}}{49\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{512}{7203}}={\left(\frac{512}{7203}\right)}^{\frac{1}{2}}$

#### Question 2:

Simplify:

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

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

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

(iv) $\left(0.001{\right)}^{1/3}$

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

(vi)

(vii)

(1) Given

By using law of rational exponents we have

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

$=\sqrt[5]{{\left(\frac{1}{32}\right)}^{3}}\phantom{\rule{0ex}{0ex}}={\left(\frac{1}{32}\right)}^{\frac{3}{5}}\phantom{\rule{0ex}{0ex}}={\left(\frac{1}{{\left(2\right)}^{5}}\right)}^{\frac{3}{5}}\phantom{\rule{0ex}{0ex}}={\left(\frac{1}{2}\right)}^{5×\frac{3}{5}}$
$={\left(\frac{1}{2}\right)}^{3}\phantom{\rule{0ex}{0ex}}=\left(\frac{1}{8}\right)$

(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

#### Question 3:

Prove that:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

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

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

(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,

#### Question 4:

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

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

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

(iv) ${\left(\frac{{x}^{{a}^{2}+{b}^{2}}}{{x}^{ab}}\right)}^{a+b}{\left(\frac{{x}^{{b}^{2}+{c}^{2}}}{{x}^{bc}}\right)}^{b+c}{\left(\frac{{x}^{{c}^{2}+{a}^{2}}}{{x}^{ac}}\right)}^{a+c}={x}^{2\left({a}^{3}+{b}^{3}+{c}^{3}\right)}$

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

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

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

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

(i)
$\frac{1}{1+{x}^{a-b}}+\frac{1}{1+{x}^{b-a}}$
$=\frac{1}{1+\frac{{x}^{a}}{{x}^{b}}}+\frac{1}{1+\frac{{x}^{b}}{{x}^{a}}}\phantom{\rule{0ex}{0ex}}=\frac{{x}^{b}}{{x}^{b}+{x}^{a}}+\frac{{x}^{a}}{{x}^{a}+{x}^{b}}\phantom{\rule{0ex}{0ex}}=\frac{{x}^{b}+{x}^{a}}{{x}^{a}+{x}^{b}}\phantom{\rule{0ex}{0ex}}=1$

(ii)
${\left[\left\{\frac{{x}^{a\left(a-b\right)}}{{x}^{a\left(a+b\right)}}\right\}÷\left\{\frac{{x}^{b\left(b-a\right)}}{{x}^{b\left(b+a\right)}}\right\}\right]}^{a+b}\phantom{\rule{0ex}{0ex}}={\left[\left\{\frac{{x}^{a\left(a-b\right)}}{{x}^{a\left(a+b\right)}}\right\}×\left\{\frac{{x}^{b\left(b+a\right)}}{{x}^{b\left(b-a\right)}}\right\}\right]}^{a+b}\phantom{\rule{0ex}{0ex}}={\left[\left\{\frac{{x}^{{a}^{2}-ab}}{{x}^{{a}^{2}+ab}}\right\}×\left\{\frac{{x}^{{b}^{2}+ab}}{{x}^{{b}^{2}-ab}}\right\}\right]}^{a+b}$
$={\left[\left\{{x}^{{a}^{2}-ab-{a}^{2}-ab}\right\}×\left\{{x}^{{b}^{2}+ab-{b}^{2}+ab}\right\}\right]}^{a+b}\phantom{\rule{0ex}{0ex}}={\left[{x}^{-2ab}×{x}^{2ab}\right]}^{a+b}\phantom{\rule{0ex}{0ex}}={\left[{x}^{-2ab+2ab}\right]}^{a+b}\phantom{\rule{0ex}{0ex}}={\left[{x}^{0}\right]}^{a+b}\phantom{\rule{0ex}{0ex}}=1$

(iii)
${\left({x}^{\frac{1}{a-b}}\right)}^{\frac{1}{a-c}}{\left({x}^{\frac{1}{b-c}}\right)}^{\frac{1}{b-a}}{\left({x}^{\frac{1}{c-a}}\right)}^{\frac{1}{c-b}}\phantom{\rule{0ex}{0ex}}={\left(x\right)}^{\frac{1}{a-b}×\frac{1}{a-c}}{{\left(x\right)}^{\frac{1}{b-c}×}}^{\frac{1}{b-a}}{{\left(x\right)}^{\frac{1}{c-a}×}}^{\frac{1}{c-b}}\phantom{\rule{0ex}{0ex}}={\left(x\right)}^{\frac{1}{a-b}×\frac{1}{a-c}+\frac{1}{b-c}×\frac{1}{b-a}+\frac{1}{c-a}×\frac{1}{c-b}}\phantom{\rule{0ex}{0ex}}={\left(x\right)}^{\frac{\left(b-c\right)-\left(a-c\right)+\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}}\phantom{\rule{0ex}{0ex}}={x}^{0}\phantom{\rule{0ex}{0ex}}=1$

(iv)
${\left(\frac{{x}^{{a}^{2}+{b}^{2}}}{{x}^{ab}}\right)}^{a+b}{\left(\frac{{x}^{{b}^{2}+{c}^{2}}}{{x}^{bc}}\right)}^{b+c}{\left(\frac{{x}^{{c}^{2}+{a}^{2}}}{{x}^{ac}}\right)}^{a+c}\phantom{\rule{0ex}{0ex}}={\left({x}^{{a}^{2}+{b}^{2}-ab}\right)}^{a+b}{\left({x}^{{b}^{2}+{c}^{2}-bc}\right)}^{b+c}{\left({x}^{{c}^{2}+{a}^{2}-ac}\right)}^{a+c}\phantom{\rule{0ex}{0ex}}=\left[{x}^{\left(a+b\right)\left({a}^{2}+{b}^{2}-ab\right)}\right]\left[{x}^{\left(b+c\right)\left({b}^{2}+{c}^{2}-bc\right)}\right]\left[{x}^{\left(a+c\right)\left({c}^{2}+{a}^{2}-ac\right)}\right]\phantom{\rule{0ex}{0ex}}=\left({x}^{{a}^{3}+{b}^{3}}\right)\left({x}^{{b}^{3}+{c}^{3}}\right)\left({x}^{{a}^{3}+{c}^{3}}\right)\phantom{\rule{0ex}{0ex}}={x}^{2\left({a}^{3}+{b}^{3}+{c}^{3}\right)}$

(v)
${\left({x}^{a-b}\right)}^{a+b}{\left({x}^{b-c}\right)}^{b+c}{\left({x}^{c-a}\right)}^{c+a}\phantom{\rule{0ex}{0ex}}=\left[{x}^{\left(a-b\right)\left(a+b\right)}\right]\left[{x}^{\left(b-c\right)\left(b+c\right)}\right]\left[{x}^{\left(c-a\right)\left(c+a\right)}\right]\phantom{\rule{0ex}{0ex}}={x}^{\left({a}^{2}-{b}^{2}\right)}{x}^{\left({b}^{2}-{c}^{2}\right)}{x}^{\left({c}^{2}-{a}^{2}\right)}\phantom{\rule{0ex}{0ex}}={x}^{{a}^{2}-{b}^{2}+{b}^{2}-{c}^{2}+{c}^{2}-{a}^{2}}\phantom{\rule{0ex}{0ex}}={x}^{0}\phantom{\rule{0ex}{0ex}}=1$

(vi)
${\left\{{\left({x}^{a-{a}^{-1}}\right)}^{\frac{1}{a-1}}\right\}}^{\frac{a}{a+1}}\phantom{\rule{0ex}{0ex}}=\left\{{\left({x}^{a-\frac{1}{a}}\right)}^{\frac{1}{a-1}×\frac{a}{a+1}}\right\}\phantom{\rule{0ex}{0ex}}={\left\{{x}^{\frac{{a}^{2}-1}{a}}\right\}}^{\frac{a}{{a}^{2}-1}}\phantom{\rule{0ex}{0ex}}={x}^{\frac{{a}^{2}-1}{a}×\frac{a}{{a}^{2}-1}}\phantom{\rule{0ex}{0ex}}={x}^{1}\phantom{\rule{0ex}{0ex}}=x$

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

(viii)
${\left(\frac{{3}^{a}}{{3}^{b}}\right)}^{a+b}{\left(\frac{{3}^{b}}{{3}^{c}}\right)}^{b+c}{\left(\frac{{3}^{c}}{{3}^{a}}\right)}^{c+a}\phantom{\rule{0ex}{0ex}}={\left({3}^{a-b}\right)}^{a+b}{\left({3}^{b-c}\right)}^{b+c}{\left({3}^{c-a}\right)}^{c+a}\phantom{\rule{0ex}{0ex}}={3}^{{a}^{2}-{b}^{2}+{b}^{2}-{c}^{2}+{c}^{2}-{a}^{2}}\phantom{\rule{0ex}{0ex}}={3}^{0}\phantom{\rule{0ex}{0ex}}=1$

#### Question 5:

If .

Let ${2}^{x}={3}^{y}={12}^{z}=k$
$⇒2={k}^{\frac{1}{x}},3={k}^{\frac{1}{y}},12={k}^{\frac{1}{z}}$
Now,
$12={k}^{\frac{1}{z}}\phantom{\rule{0ex}{0ex}}⇒{2}^{2}×3={k}^{\frac{1}{z}}\phantom{\rule{0ex}{0ex}}⇒{\left({k}^{\frac{1}{x}}\right)}^{2}×{k}^{\frac{1}{y}}={k}^{\frac{1}{z}}\phantom{\rule{0ex}{0ex}}⇒{k}^{\frac{2}{x}+\frac{1}{y}}={k}^{\frac{1}{z}}\phantom{\rule{0ex}{0ex}}⇒\frac{2}{x}+\frac{1}{y}=\frac{1}{z}$

#### Question 6:

If .

Let ${2}^{x}={3}^{y}={6}^{-z}=k$

Now,
$6=2×3={k}^{\frac{1}{-z}}\phantom{\rule{0ex}{0ex}}⇒{k}^{\frac{1}{x}}×{k}^{\frac{1}{y}}={k}^{\frac{1}{-z}}\phantom{\rule{0ex}{0ex}}⇒{k}^{\frac{1}{x}+\frac{1}{y}}={k}^{\frac{1}{-z}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{x}+\frac{1}{y}=\frac{1}{-z}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$

#### Question 7:

If ax = by = cz and b2 = ac, then show that $y=\frac{2zx}{z+x}$.

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,
${b}^{2}=ac\phantom{\rule{0ex}{0ex}}⇒{\left({k}^{\frac{1}{y}}\right)}^{2}=\left({k}^{\frac{1}{x}}\right)\left({k}^{\frac{1}{z}}\right)\phantom{\rule{0ex}{0ex}}⇒{k}^{\frac{2}{y}}={k}^{\frac{1}{x}+\frac{1}{z}}\phantom{\rule{0ex}{0ex}}⇒\frac{2}{y}=\frac{1}{x}+\frac{1}{z}$
$⇒\frac{2}{y}=\frac{z+x}{xz}\phantom{\rule{0ex}{0ex}}⇒y=\frac{2zx}{z+x}$

#### Question 8:

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

Let ${3}^{x}={5}^{y}={\left(75\right)}^{z}=k$

$⇒{k}^{\frac{2}{y}+\frac{1}{x}}={k}^{\frac{1}{z}}\phantom{\rule{0ex}{0ex}}⇒\frac{2}{y}+\frac{1}{x}=\frac{1}{z}\phantom{\rule{0ex}{0ex}}⇒\frac{2x+y}{xy}=\frac{1}{z}\phantom{\rule{0ex}{0ex}}⇒z=\frac{xy}{2x+y}$

#### Question 9:

If find x.

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,

#### Question 10:

Find the values of x in each of the following:

(i)

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

(iii)

(iv)

(v) ${2}^{x-7}×{5}^{x-4}=1250$

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

(vii) ${5}^{2x+3}=1$

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

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

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

${\left({2}^{2}\right)}^{\frac{1}{3}\left(\frac{4x+1}{2}\right)}={\left(\frac{1}{2}\right)}^{5}\phantom{\rule{0ex}{0ex}}⇒{2}^{\left(\frac{4x+1}{3}\right)}={2}^{-5}$

On comparing we get,
$\frac{4x+1}{3}=-5\phantom{\rule{0ex}{0ex}}⇒4x+1=-15\phantom{\rule{0ex}{0ex}}⇒4x=-16\phantom{\rule{0ex}{0ex}}⇒x=-4$

(vii) ${5}^{2x+3}=1$

${5}^{2x+3}={5}^{0}\phantom{\rule{0ex}{0ex}}⇒2x+3=0\phantom{\rule{0ex}{0ex}}⇒x=\frac{-3}{2}$

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

${\left(13\right)}^{\sqrt{x}}={\left({2}^{2}\right)}^{4}-{3}^{4}-6\phantom{\rule{0ex}{0ex}}⇒{\left(13\right)}^{\sqrt{x}}={2}^{8}-{3}^{4}-6\phantom{\rule{0ex}{0ex}}⇒{\left(13\right)}^{\sqrt{x}}=256-81-6\phantom{\rule{0ex}{0ex}}⇒{\left(13\right)}^{\sqrt{x}}=169\phantom{\rule{0ex}{0ex}}⇒{\left(13\right)}^{\sqrt{x}}={\left(13\right)}^{2}$

On comparing we get,

(ix) ${\left(\sqrt{\frac{3}{5}}\right)}^{x+1}=\frac{125}{27}$
${\left(\sqrt{\frac{3}{5}}\right)}^{x+1}={\left(\frac{5}{3}\right)}^{3}\phantom{\rule{0ex}{0ex}}⇒{\left(\frac{3}{5}\right)}^{\frac{x+1}{2}}={\left(\frac{3}{5}\right)}^{-3}$

On comparing we get,

$\frac{x+1}{2}=-3\phantom{\rule{0ex}{0ex}}⇒x+1=-6\phantom{\rule{0ex}{0ex}}⇒x=-7$

#### Question 11:

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

$x={2}^{\frac{1}{3}}+{2}^{\frac{2}{3}}$
Cubing on both sides, we get
${x}^{3}={\left({2}^{\frac{1}{3}}+{2}^{\frac{2}{3}}\right)}^{3}\phantom{\rule{0ex}{0ex}}⇒{x}^{3}={\left({2}^{\frac{1}{3}}\right)}^{3}+{\left({2}^{\frac{2}{3}}\right)}^{3}+3×{2}^{\frac{1}{3}}×{2}^{\frac{2}{3}}\left({2}^{\frac{1}{3}}+{2}^{\frac{2}{3}}\right)\phantom{\rule{0ex}{0ex}}⇒{x}^{3}=2+{2}^{2}+3×{2}^{\frac{1+2}{3}}\left(x\right)\phantom{\rule{0ex}{0ex}}⇒{x}^{3}=2+4+3×2x\phantom{\rule{0ex}{0ex}}⇒{x}^{3}-6x=6$

#### Question 12:

Determine  .

${9}^{x+2}=240+{9}^{x}$
$⇒{9}^{x}×{9}^{2}=240+{9}^{x}\phantom{\rule{0ex}{0ex}}⇒{9}^{x}\left({9}^{2}-1\right)=240\phantom{\rule{0ex}{0ex}}⇒{9}^{x}\left(81-1\right)=240\phantom{\rule{0ex}{0ex}}⇒{9}^{x}×80=240\phantom{\rule{0ex}{0ex}}⇒{9}^{x}=3\phantom{\rule{0ex}{0ex}}⇒{3}^{2x}={3}^{1}\phantom{\rule{0ex}{0ex}}⇒2x=1\phantom{\rule{0ex}{0ex}}⇒x=\frac{1}{2}$
$\therefore {\left(8x\right)}^{x}={\left(8×\frac{1}{2}\right)}^{\frac{1}{2}}={\left(4\right)}^{\frac{1}{2}}=2$

#### Question 13:

If ${3}^{x+1}={9}^{x-2}$, find the value of 21+x.

${3}^{x+1}={9}^{x-2}$
$⇒{3}^{x}×3=\frac{{9}^{x}}{{9}^{2}}\phantom{\rule{0ex}{0ex}}⇒{3}^{x}×3=\frac{{\left({3}^{2}\right)}^{x}}{{\left({3}^{2}\right)}^{2}}=\frac{{3}^{2x}}{{3}^{4}}\phantom{\rule{0ex}{0ex}}⇒{3}^{4}×3=\frac{{3}^{2x}}{{3}^{x}}\phantom{\rule{0ex}{0ex}}⇒{3}^{5}={3}^{2x-x}$
$⇒{3}^{5}={3}^{x}$
Comparing both sides, we get
x = 5
So, ${2}^{1+x}={2}^{1+5}={2}^{6}=64$

#### Question 14:

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

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

#### Question 15:

If , find x and y.

It is given that .
Now,
${5}^{3x}=125\phantom{\rule{0ex}{0ex}}⇒{5}^{3x}={5}^{3}\phantom{\rule{0ex}{0ex}}⇒3x=3\phantom{\rule{0ex}{0ex}}⇒x=1$
And,
${10}^{y}=0.001\phantom{\rule{0ex}{0ex}}⇒{10}^{y}=\frac{1}{1000}\phantom{\rule{0ex}{0ex}}⇒{10}^{y}={10}^{-3}\phantom{\rule{0ex}{0ex}}⇒y=-3$
Hence, the values of x and y are 1 and −3, respectively.

#### Question 16:

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

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

(i)
${3}^{x+1}=27×{3}^{4}\phantom{\rule{0ex}{0ex}}⇒{3}^{x+1}={3}^{3}×{3}^{4}\phantom{\rule{0ex}{0ex}}⇒{3}^{x+1}={3}^{3+4}\phantom{\rule{0ex}{0ex}}⇒{3}^{x+1}={3}^{7}\phantom{\rule{0ex}{0ex}}⇒x+1=7\phantom{\rule{0ex}{0ex}}⇒x=6$

(ii)

(iii)

(iv)

Now,
$3+x=6y⇒x=6y-3$
Putting x = 6y − 3 in $3x-4y=5$, we get
$3\left(6y-3\right)-4y=5\phantom{\rule{0ex}{0ex}}⇒18y-9-4y=5\phantom{\rule{0ex}{0ex}}⇒14y=14\phantom{\rule{0ex}{0ex}}⇒y=1$
Putting y = 1 in $x=6y-3$, we get
$x=6×1-3=3$

(v)
${4}^{x-1}×{\left(0.5\right)}^{3-2x}={\left(\frac{1}{8}\right)}^{x}\phantom{\rule{0ex}{0ex}}⇒{\left({2}^{2}\right)}^{x-1}×{\left(\frac{1}{2}\right)}^{3-2x}={\left(\frac{1}{{2}^{3}}\right)}^{x}\phantom{\rule{0ex}{0ex}}⇒{\left(2\right)}^{2x-2}×{\left(2\right)}^{-\left(3-2x\right)}={\left(2\right)}^{-3x}\phantom{\rule{0ex}{0ex}}⇒{\left(2\right)}^{2x-2-3+2x}={\left(2\right)}^{-3x}\phantom{\rule{0ex}{0ex}}⇒4x-5=-3x\phantom{\rule{0ex}{0ex}}⇒7x=5\phantom{\rule{0ex}{0ex}}⇒x=\frac{5}{7}$

(vi)
$\sqrt{\frac{a}{b}}={\left(\frac{b}{a}\right)}^{1-2x}\phantom{\rule{0ex}{0ex}}⇒{\left(\frac{a}{b}\right)}^{\frac{1}{2}}={\left(\frac{a}{b}\right)}^{-\left(1-2x\right)}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}=-\left(1-2x\right)\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}=2x-1\phantom{\rule{0ex}{0ex}}⇒\frac{3}{2}=2x\phantom{\rule{0ex}{0ex}}⇒x=\frac{3}{4}$

#### Question 17:

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

Given: $\sqrt[3]{{a}^{6}{b}^{-4}}={a}^{x}{b}^{2y}$

#### Question 18:

If a and b are different positive primes such that

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

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

(i)

(ii)

Therefore, the value of x + y + 2 is −1 −1 + 2 = 0.

#### Question 19:

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

Given: ${2}^{x}×{3}^{y}×{5}^{z}=2160$
First, find out the prime factorisation of 2160.
$\begin{array}{cc}2& 2160\\ 2& 1080\\ 2& 540\\ 2& 270\\ 3& 135\\ 3& 45\\ 3& 15\\ 5& 5\\ & 1\end{array}$
It can be observed that 2160 can be written as ${2}^{4}×{3}^{3}×{5}^{1}$.
Also,

Therefore, the value of ${3}^{x}×{2}^{-y}×{5}^{-z}$ is ${3}^{4}×{2}^{-3}×{5}^{-1}=81×\frac{1}{8}×\frac{1}{5}=\frac{81}{40}$.

#### Question 20:

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

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

#### Question 21:

Simplify:

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

(ii) $lm\sqrt{\frac{{x}^{l}}{{x}^{m}}}×mn\sqrt{\frac{{x}^{m}}{{x}^{n}}}×nl\sqrt{\frac{{x}^{n}}{{x}^{l}}}$

(i)
${\left(\frac{{x}^{a+b}}{{x}^{c}}\right)}^{a-b}{\left(\frac{{x}^{b+c}}{{x}^{a}}\right)}^{b-c}{\left(\frac{{x}^{c+a}}{{x}^{b}}\right)}^{c-a}\phantom{\rule{0ex}{0ex}}=\left(\frac{{x}^{\left(a+b\right)\left(a-b\right)}}{{x}^{c\left(a-b\right)}}\right)\left(\frac{{x}^{\left(b+c\right)\left(b-c\right)}}{{x}^{a\left(b-c\right)}}\right)\left(\frac{{x}^{\left(c+a\right)\left(c-a\right)}}{{x}^{b\left(c-a\right)}}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{{x}^{\left(a+b\right)\left(a-b\right)}}{{x}^{ca-bc}}\right)\left(\frac{{x}^{\left(b+c\right)\left(b-c\right)}}{{x}^{ab-ac}}\right)\left(\frac{{x}^{\left(c+a\right)\left(c-a\right)}}{{x}^{bc-ab}}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{{x}^{\left({a}^{2}-{b}^{2}\right)}{x}^{\left({b}^{2}-{c}^{2}\right)}{x}^{\left({c}^{2}-{a}^{2}\right)}}{{x}^{ca-bc+ab-ac+bc-ab}}\right)\phantom{\rule{0ex}{0ex}}=\frac{{x}^{{a}^{2}-{b}^{2}+{b}^{2}-{c}^{2}+{c}^{2}-{a}^{2}}}{{x}^{ca-bc+ab-ac+bc-ab}}\phantom{\rule{0ex}{0ex}}=\frac{{x}^{0}}{{x}^{0}}\phantom{\rule{0ex}{0ex}}=1\phantom{\rule{0ex}{0ex}}$

(ii)
$\sqrt[lm]{\frac{{x}^{l}}{{x}^{m}}}×\sqrt[mn]{\frac{{x}^{m}}{{x}^{n}}}×\sqrt[nl]{\frac{{x}^{n}}{{x}^{l}}}\phantom{\rule{0ex}{0ex}}={\left(\frac{{x}^{l}}{{x}^{m}}\right)}^{\frac{1}{ml}}×{\left(\frac{{x}^{m}}{{x}^{n}}\right)}^{\frac{1}{mn}}×{\left(\frac{{x}^{n}}{{x}^{l}}\right)}^{\frac{1}{nl}}\phantom{\rule{0ex}{0ex}}={\left({x}^{l-m}\right)}^{\frac{1}{ml}}×{\left({x}^{m-n}\right)}^{\frac{1}{mn}}×{\left({x}^{n-l}\right)}^{\frac{1}{nl}}\phantom{\rule{0ex}{0ex}}={x}^{\frac{l-m}{ml}}×{x}^{\frac{m-n}{mn}}×{x}^{\frac{n-l}{nl}}\phantom{\rule{0ex}{0ex}}={x}^{\frac{l-m}{ml}+\frac{m-n}{mn}+\frac{n-l}{nl}}\phantom{\rule{0ex}{0ex}}={x}^{\frac{ln-mn+lm-nl+nm-lm}{nml}}\phantom{\rule{0ex}{0ex}}={x}^{0}\phantom{\rule{0ex}{0ex}}=1\phantom{\rule{0ex}{0ex}}$

#### Question 22:

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

$\frac{{\left(a+\frac{1}{b}\right)}^{m}×{\left(a-\frac{1}{b}\right)}^{n}}{{\left(b+\frac{1}{a}\right)}^{m}×{\left(b-\frac{1}{a}\right)}^{n}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(\frac{ab+1}{b}\right)}^{m}×{\left(\frac{ab-1}{b}\right)}^{n}}{{\left(\frac{ab+1}{a}\right)}^{m}×{\left(\frac{ab-1}{a}\right)}^{n}}\phantom{\rule{0ex}{0ex}}={\left(\frac{\frac{ab+1}{b}}{\frac{ab+1}{a}}\right)}^{m}×{\left(\frac{\frac{ab-1}{b}}{\frac{ab-1}{a}}\right)}^{n}\phantom{\rule{0ex}{0ex}}={\left(\frac{ab+1}{b}×\frac{a}{ab+1}\right)}^{m}×{\left(\frac{ab-1}{b}×\frac{a}{ab-1}\right)}^{n}\phantom{\rule{0ex}{0ex}}={\left(\frac{a}{b}\right)}^{m}×{\left(\frac{a}{b}\right)}^{n}\phantom{\rule{0ex}{0ex}}={\left(\frac{a}{b}\right)}^{m}×{\left(\frac{a}{b}\right)}^{n}\phantom{\rule{0ex}{0ex}}={\left(\frac{a}{b}\right)}^{m+n}$

#### Question 23:

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

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

(i) Given:
Putting the values of a, b and c in ${a}^{m-n}{b}^{n-l}{c}^{l-m}$, we get
${a}^{m-n}{b}^{n-l}{c}^{l-m}\phantom{\rule{0ex}{0ex}}={\left({x}^{m+n}{y}^{l}\right)}^{m-n}{\left({x}^{n+l}{y}^{m}\right)}^{n-l}{\left({x}^{l+m}{y}^{n}\right)}^{l-m}\phantom{\rule{0ex}{0ex}}=\left[{x}^{\left(m+n\right)\left(m-n\right)}{y}^{l\left(m-n\right)}\right]\left[{x}^{\left(n+l\right)\left(n-l\right)}{y}^{m\left(n-l\right)}\right]\left[{x}^{\left(l+m\right)\left(l-m\right)}{y}^{n\left(l-m\right)}\right]\phantom{\rule{0ex}{0ex}}={x}^{\left({m}^{2}-{n}^{2}\right)}{x}^{\left({n}^{2}-{l}^{2}\right)}{x}^{\left({l}^{2}-{m}^{2}\right)}{y}^{lm-ln}{y}^{mn-ml}{y}^{nl-nm}\phantom{\rule{0ex}{0ex}}={x}^{{m}^{2}-{n}^{2}+{n}^{2}-{l}^{2}+{l}^{2}-{m}^{2}}{y}^{lm-ln+mn-ml+nl-nm}\phantom{\rule{0ex}{0ex}}={x}^{0}{y}^{0}\phantom{\rule{0ex}{0ex}}=1$

(ii) Given:
Putting the values of x, y and z in ${x}^{m}{y}^{n}{z}^{l}$, we get
${x}^{m}{y}^{n}{z}^{l}\phantom{\rule{0ex}{0ex}}={\left({a}^{m+n}\right)}^{m}{\left({a}^{n+l}\right)}^{n}{\left({a}^{l+m}\right)}^{l}\phantom{\rule{0ex}{0ex}}=\left({a}^{{m}^{2}+nm}\right)\left({a}^{{n}^{2}+\mathrm{l}n}\right)\left({a}^{{l}^{2}+lm}\right)\phantom{\rule{0ex}{0ex}}={a}^{{m}^{2}+{n}^{2}+{l}^{2}+nm+ln+lm}$
Putting the values of x, y and z in ${x}^{n}{y}^{l}{z}^{m}$, we get

So, ${x}^{m}{y}^{n}{z}^{l}$ = ${x}^{n}{y}^{l}{z}^{m}$

#### Question 1:

The value of is

(a) 5

(b) 125

(c) 1/5

(d) -125

We have to find the value of. So,

The value of is 125

Hence the correct choice is

#### Question 2:

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

(a) 18
(b) −18
(c) 14
(d) −14

Given

Here

By substituting in we get

The value of is – 14

Hence the correct choice is .

#### 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

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

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

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 .

#### Question 5:

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

(a) 64
(b) 2
(c) $\frac{1}{2}$
(d) 642/3

We have to find the value of

So,

The value of is

Hence the correct choice is .

#### Question 6:

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

(a) 400

(b) 324

(c) 289

(d) 196

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

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

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

Hence the correct answer is option (a).

#### Question 7:

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

(a) xy

(b) x+y

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

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

We have to simplify

So,

The value of is

Hence the correct choice is .

#### Question 8:

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

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

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 .

#### Question 9:

If (23)2 = 4x, then 3x =

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

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

#### Question 10:

If x-2 = 64, then x1/3+x0 =

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

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 .

#### Question 11:

When simplified is

(a) 9

(b) −9

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

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

We have to find the value of

So,

Hence the correct choice is .

#### Question 12:

Which one of the following is not equal to

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

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

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

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

We have to find the value of

So,

Also,

Hence the correct alternative is .

#### Question 13:

Which one of the following is not equal to ?

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

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

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

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

We have to find the value of

So,

Since, is equal to ,,.

Hence the correct choice is

#### Question 14:

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

(a) 1

(b) abc

(c) $\sqrt{abc}$

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

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

So,

Taking square root as common we get

$\sqrt{{a}^{-1}b}×\sqrt{{b}^{-1}c}×\sqrt{{c}^{-1}a}=\sqrt{\frac{b}{a}×\frac{c}{b}×\frac{a}{c}}\phantom{\rule{0ex}{0ex}}\sqrt{{a}^{-1}b}×\sqrt{{b}^{-1}c}×\sqrt{{c}^{-1}a}=1$

Hence the correct alternative is .

#### Question 15:

, then x =

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

We have to find value of provided

So,

Equating exponents of power we get

Hence the correct alternative is

#### Question 16:

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

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

(b) 2

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

(d) 4

Find the value of

Hence the correct choice is .

#### 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

Find value of.

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

Hence the correct choice is .

#### Question 18:

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

(a) amn

(b) a

(c) am/n

(d) 1

Find the value of .

So,

Hence the correct choice is

#### Question 19:

If x = 2 and y = 4, then

(a) 4

(b) 8

(c) 12

(d) 2

We have to find the value of if,

Substitute,into get,

Hence the correct choice is .

#### Question 20:

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

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

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

(c) −3

(d) 2

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 .

#### 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

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

Hence, the correct answer is option (d).

#### Question 22:

(256)0.16 × (256)0.09

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

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 .

#### Question 23:

If 102y = 25, then 10-y equals

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

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

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

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

We have to find the value of

Given that, therefore,

Hence the correct option is .

#### Question 24:

If 9x+2 = 240 + 9x, then x =

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

We have to find the value of

Given

By equating the exponents we get

Hence the correct alternative is .

#### 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

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 .

#### 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

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

#### Question 27:

If , What is the value of g when t = 64?

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

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

(c) $16$

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

Given.We have to find the value of

So,

The value of is

Hence the correct choice is

#### Question 28:

If then (2x)x equals

(a) $5\sqrt{5}$

(b) $\sqrt{5}$

(c) $25\sqrt{5}$

(d) 125

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

#### Question 29:

When simplified  is

(a) 8

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

(c) 2

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

Simplify

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

Hence the correct choice is .

#### Question 30:

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

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

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

#### Question 31:

The value of 64-1/3 (641/3-642/3), is

(a) 1

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

(c) −3

(d) −2

Find the value of

So,

Hence the correct statement is.

#### Question 32:

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

(a) 25

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

(c) 625

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

We have to find provided

So,

Substitute in to get

Hence the value of is

The correct choice is

#### Question 33:

If (16)2x+3 =(64)x+3, then 42x-2 =

(a) 64

(b) 256

(c) 32

(d) 512

We have to find the value ofprovided

So,

Equating the power of exponents we get

The value of is

Hence the correct alternative is

#### Question 34:

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

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

(b) 2

(c) 4

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

We have to find the value ofprovided

Consider,

Equating the power of exponents we get

By substituting we get

Hence the correct choice is .

#### Question 35:

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

(a) 2

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

(c) 9

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

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

Find :
By using rational components We get

By equating rational exponents we get

Now,   =$\left({a}^{2m+n-p}\right).\left({a}^{m-2n+2p}\right)$  we get

Also, $\frac{{3}^{p}}{{3}^{n}}=81$
${3}^{p-n}={3}^{4}\phantom{\rule{0ex}{0ex}}$
On comparing LHS and RHS we get, p - n = 4.
Now,
= a3m - n + p
$={2}^{\frac{6+\left(p-n\right)}{10}}\phantom{\rule{0ex}{0ex}}={2}^{\frac{6+4}{10}}\phantom{\rule{0ex}{0ex}}={2}^{\frac{10}{10}}={2}^{1}\phantom{\rule{0ex}{0ex}}=2$

So, option (a) is the correct answer.

#### Question 36:

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

(a) 3

(b) −3

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

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

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 .

#### Question 37:

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

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 .

#### Question 38:

If 10x = 64, what is the value of

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

We have to find the value of provided

So,

By substituting we get

Hence the correct choice is .

#### Question 39:

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

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

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

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

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

We have to simplify

Taking as a common factor we get

Hence the correct alternative is

#### Question 40:

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

(a) 3

(b) 9

(c) 27

(d) 81

We have to find

Given

Equating powers of rational exponents we get

Substituting in we get

Hence the correct choice is .

#### Question 1:

(212 – 152)2/3 is equal to ________.

Hence, (212 – 152)2/3 is equal to 36.

#### Question 2:

811/4 × 93/2 × 27–4/3 is equal to _________.

Hence, 811/4 × 93/2 × 27–4/3 is equal to 1.

#### Question 3:

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

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

#### Question 4:

If x = 82/3 × 32–2/5, then x–5 = ________.

Hence, if x = 82/3 × 32–2/5, then x–5 = 1.

#### Question 5:

If 6n = 1296, then 6n–3 = _________.

Hence, if 6n = 1296, then 6n–3 = 6.

#### Question 6:

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

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

If

Hence, if

#### Question 8:

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

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

#### Question 9:

If

$\frac{{\left(p+\frac{1}{q}\right)}^{p-q}{\left(p-\frac{1}{q}\right)}^{p+q}}{{\left(q+\frac{1}{p}\right)}^{p-q}{\left(q-\frac{1}{p}\right)}^{p+q}}={\left(\frac{p}{q}\right)}^{x}\phantom{\rule{0ex}{0ex}}⇒\frac{{\left(\frac{qp+1}{q}\right)}^{p-q}{\left(\frac{pq-1}{q}\right)}^{p+q}}{{\left(\frac{pq+1}{p}\right)}^{p-q}{\left(\frac{pq-1}{p}\right)}^{p+q}}={\left(\frac{p}{q}\right)}^{x}\phantom{\rule{0ex}{0ex}}⇒\frac{\frac{{\left(qp+1\right)}^{p-q}}{{\left(q\right)}^{p-q}}×\frac{{\left(pq-1\right)}^{p+q}}{{\left(q\right)}^{p+q}}}{\frac{{\left(pq+1\right)}^{p-q}}{{\left(p\right)}^{p-q}}×\frac{{\left(pq-1\right)}^{p+q}}{{\left(p\right)}^{p+q}}}={\left(\frac{p}{q}\right)}^{x}\phantom{\rule{0ex}{0ex}}⇒\frac{{\left(qp+1\right)}^{p-q}}{{\left(q\right)}^{p-q}}×\frac{{\left(pq-1\right)}^{p+q}}{{\left(q\right)}^{p+q}}×\frac{{\left(p\right)}^{p-q}}{{\left(pq+1\right)}^{p-q}}×\frac{{\left(p\right)}^{p+q}}{{\left(pq-1\right)}^{p+q}}={\left(\frac{p}{q}\right)}^{x}\phantom{\rule{0ex}{0ex}}⇒\frac{{\left(p\right)}^{p-q}}{{\left(q\right)}^{p-q}}×\frac{{\left(p\right)}^{p+q}}{{\left(q\right)}^{p+q}}={\left(\frac{p}{q}\right)}^{x}\phantom{\rule{0ex}{0ex}}⇒\frac{{\left(p\right)}^{p-q+p+q}}{{\left(q\right)}^{p-q+p+q}}={\left(\frac{p}{q}\right)}^{x}\phantom{\rule{0ex}{0ex}}⇒\frac{{\left(p\right)}^{2p}}{{\left(q\right)}^{2p}}={\left(\frac{p}{q}\right)}^{x}\phantom{\rule{0ex}{0ex}}⇒{\left(\frac{p}{q}\right)}^{2p}={\left(\frac{p}{q}\right)}^{x}\phantom{\rule{0ex}{0ex}}⇒2p=x\phantom{\rule{0ex}{0ex}}⇒x=2p\phantom{\rule{0ex}{0ex}}$

Hence, if

#### Question 10:

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

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

#### Question 11:

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

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

#### Question 12:

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

Hence, 5x + 6y = 0.

#### Question 13:

If 6x–y = 36 and 3x+y = 729, then x2y2 = _________.

Hence, x2 – y2 = 12.

#### Question 14:

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

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

#### Question 15:

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

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

#### Question 16:

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

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

#### Question 17:

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

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

#### Question 1:

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

We have to writein decimal form. So,

${\left(625\right)}^{\frac{-1}{4}}=\frac{1}{{625}^{\frac{1}{4}}}\phantom{\rule{0ex}{0ex}}={\left(\frac{1}{\left({5}^{4}\right)}\right)}^{\frac{1}{4}}$

Hence the decimal form of is

#### Question 2:

State the product law of exponents.

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.

#### Question 3:

State the quotient law of exponents.

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

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,

#### Question 4:

State the power law of exponents.

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

factors

Hence,

#### Question 5:

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

We have to find the value of x provided

So,

By equating the exponents we get

Hence the value of x is .

#### Question 6:

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

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

#### Question 7:

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

We have to find the value of . So,

By using law rational exponents we get,

Hence the value of is

#### Question 8:

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

We have to find the value of . So,

Hence the value of the value of is .

#### Question 9:

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

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

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

Hence the value of the value of is .

#### Question 10:

For any positive real number x, find the value of

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

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 .

#### Question 11:

Write the value of ${\left\{5\left({8}^{1/3}+{27}^{1/3}{\right)}^{3}\right\}}^{1/4}.\phantom{\rule{0ex}{0ex}}$

We have to find the value of. So,

By using rational exponents we get

Hence the simplified value of is

#### Question 12:

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

We have to simplify. So,

Hence, the value of is

#### Question 13:

For any positive real number x, write the value of

${\left\{{\left({x}^{a}\right)}^{b}\right\}}^{\frac{1}{ab}}{\left\{{\left({x}^{b}\right)}^{c}\right\}}^{\frac{1}{bc}}{\left\{{\left({x}^{c}\right)}^{a}\right\}}^{\frac{1}{ca}}$

We have to simplify. So,

By using rational exponents, we get

Hence the value of is

#### Question 14:

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

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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