# 7th Class Mathematics Exponents and Power Negative Rational Number as Exponent

## Negative Rational Number as Exponent

Category : 7th Class

### Negative Rational Number as Exponent

Let us consider 'a' be a positive rational number and $x\left( i.e{{.}^{\frac{-m}{n}}} \right)$ be a negative rational exponent then it is defined as ${{a}^{x}}\left( i.e.{{a}^{\frac{-m}{n}}} \right).$

Find the value of ${{4}^{\frac{-3}{2}}}$ and ${{\left( 512 \right)}^{\frac{-2}{9}}}$

Solution:

${{4}^{\frac{-3}{2}}}=\frac{1}{^{{{4}^{\frac{3}{2}}}}}=\frac{1}{^{\left( {{4}^{3}} \right)\frac{1}{2}}}=\frac{1}{{{\left( 64 \right)}^{\frac{1}{2}}}}=\frac{1}{8}$and $\frac{1}{{{\left( 512 \right)}^{\frac{-2}{9}}}}=\frac{1}{{{\left( {{512}^{2}} \right)}^{\frac{-2}{9}}}}$$=\frac{2}{\left( {{\left\{ {{\left( {{2}^{9}} \right)}^{2}} \right\}}^{\frac{1}{9}}} \right)}=\frac{1}{^{_{2}\cancel{9}\times 2\times \frac{1}{\cancel{9}}}}=\frac{1}{4}$

Evaluate : ${{\left( \frac{4}{9} \right)}^{\frac{3}{2}}}\times {{\left( \frac{4}{9} \right)}^{\frac{1}{2}}}$

(a) ${{\left( \frac{4}{9} \right)}^{2}}$

(b) ${{\left( \frac{81}{16} \right)}^{4}}$

(c) ${{\left( \frac{4}{9} \right)}^{4}}$

(d) $\frac{4}{27}$

(e) None of these

Solution:

${{\left( \frac{4}{9} \right)}^{\frac{3}{2}}}\times {{\left( \frac{4}{9} \right)}^{\frac{1}{2}}}={{\left( \frac{4}{9} \right)}^{\frac{3}{2}+\frac{1}{2}}}={{\left( \frac{4}{9} \right)}^{\frac{3+1}{2}}}={{\left( \frac{4}{9} \right)}^{\frac{\cancel{4}}{\cancel{2}}}}={{\left( \frac{4}{9} \right)}^{2}}$

Evaluate: ${{(27)}^{\frac{6}{5}}}\div {{(27)}^{\frac{1}{5}}}$

(a) $5\times {{3}^{3}}$

(b) $1\times {{3}^{3}}$

(c) $2\times {{3}^{3}}$

(d) $7\times {{3}^{3}}$

(e) None of these

Solution:

${{(27)}^{\frac{6}{5}}}\div {{(27)}^{\frac{1}{5}}}={{(27)}^{\frac{6}{5}-\frac{1}{5}}}={{(27)}^{\frac{5}{5}}}=27={{3}^{3}}$

Evaluate: $\frac{{{\left( 27 \right)}^{\frac{-2}{3}}}\times {{\left( 81 \right)}^{\frac{5}{4}}}}{{{\left( \frac{1}{3} \right)}^{-3}}}$

(a) 1

(b) 2

(c) 3

(d) 4

(e) None of these

Solution:

$\frac{{{\left( 27 \right)}^{\frac{-2}{3}}}\times {{\left( 81 \right)}^{\frac{5}{4}}}}{{{\left( \frac{1}{3} \right)}^{-3}}}=\frac{{{3}^{\cancel{3}\times \left( \frac{2}{\cancel{3}} \right)}}\times {{3}^{\cancel{4}\times \frac{5}{\cancel{4}}}}}{{{3}^{3}}}=\frac{{{3}^{-2}}\times {{3}^{5}}}{{{3}^{3}}}=\frac{{{3}^{-2+5}}}{{{3}^{3}}}=\frac{{{3}^{3}}}{{{3}^{3}}}=1$

Evaluate:  ${{\left[ {{\left( \frac{36}{25} \right)}^{\frac{3}{2}}} \right]}^{\frac{5}{3}}}$

(a) $\frac{7776}{3125}$

(b) 1

(c) $\frac{75}{31}$

(d) 2

(e) None of these

Solution:

${{\left( \frac{36}{25} \right)}^{\frac{\cancel{3}}{2}\times \frac{5}{\cancel{3}}}}{{\left( \frac{36}{25} \right)}^{\frac{5}{2}}}={{\left( \frac{6}{2} \right)}^{\cancel{2}\times \frac{5}{\cancel{2}}}}{{\left( \frac{6}{2} \right)}^{5}}=\frac{7776}{3125}$

Evaluate:  ${{\left[ {{\left\{ {{\left( 625 \right)}^{-\frac{1}{2}}} \right\}}^{\frac{1}{4}}} \right]}^{2}}$

(a) 2

(b) 3

(c) 4

(d) 5

(e) None of these

Solution

${{\left( 625 \right)}^{-\frac{1}{\cancel{2}}\times \left( -\frac{1}{4} \right)\times \cancel{2}}}={{\left( 625 \right)}^{\frac{1}{4}}}={{\left( 5 \right)}^{\cancel{4}\times \frac{1}{4}=5}}$

Evaluate: ${{64}^{\frac{2}{3}}}\times {{27}^{\frac{2}{3}}}$

(a) 144

(b) 12

(c) 4

(d) 3

(e) None of these

Solution:

${{\left( 64\times 27 \right)}^{\frac{2}{3}}}={{\left( {{4}^{3}}\times {{3}^{3}} \right)}^{\frac{2}{3}}}={{\left( 4\times 3 \right)}^{\cancel{3}\times \frac{2}{\cancel{3}}}}=16\times 9=144$

Evaluate: ${{\left[ {{\left\{ {{\left( \frac{1}{x} \right)}^{-12}} \right\}}^{\frac{1}{4}}} \right]}^{-\frac{2}{3}}}$

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

(b) $\frac{1}{{{x}^{2}}}$

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

(d)$\frac{1}{{{x}^{4}}}$

(e) None of these

Evaluate: ${{\left[ {{\left( 729 \right)}^{\frac{-5}{3}}} \right]}^{-\frac{1}{2}}}$

(a) 243

(b) 81

(c) 27

(d) 9

(e) None of these

Solution:

${{\left( 729 \right)}^{\frac{-5}{3}\times \left( \frac{-1}{2} \right)}}={{\left( 9 \right)}^{^{\cancel{3}\times \left( \frac{-5}{\cancel{3}} \right)\left( \frac{-1}{2} \right)}}}={{9}^{\frac{5}{2}}}={{3}^{\cancel{2}{{\times }^{\frac{5}{\cancel{2}}}}}}={{3}^{5}}=243$

Find the value of$x$, if ${{\left( \sqrt{6} \right)}^{x-2}}=1.$

(a) 1

(b) 2

(c) 3

(d) 4

(e) None of these

Solution:

We can write the above expression as, ${{\left( \sqrt{6} \right)}^{x-2}}={{\left( \sqrt{6} \right)}^{0}}\Rightarrow x-2=0\Rightarrow x=2$

Find the value of $x$ so that ${{2}^{2x+1}}={{4}^{2x-1}}$

(a) 1

(b) 2

(c) $\frac{3}{2}$

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

(e) None of these

Solution:

We have ${{2}^{x+1}}={{2}^{2\left( 2x-1 \right)}}\Leftrightarrow {{2}^{2x+1}}={{2}^{4x-2}}$ $\therefore 2x+1=4x-2\Rightarrow 2x-4=-2-1$ $\Rightarrow -2x=-3\Rightarrow 2x=3\Rightarrow x=\frac{3}{2}$

The value of the expression ${{\left( \frac{64}{125} \right)}^{\frac{2}{3}}}\times {{\left( \frac{64}{125} \right)}^{\frac{5}{3}}}$  is equal to:

(a) $\frac{125}{64}$

(b) $\frac{{{4}^{2}}}{{{5}^{5}}}$

(c) ${{\left( \frac{4}{5} \right)}^{7}}$

(d) $\frac{16}{25}$

(e) None of these

Solution:

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

$={{\left( \frac{64}{125} \right)}^{\frac{7}{3}}}={{\left( \frac{{{\left( 4 \right)}^{3}}}{{{\left( 5 \right)}^{3}}} \right)}^{\frac{7}{3}}}={{\left( {{\left( \frac{4}{5} \right)}^{3}} \right)}^{\frac{7}{3}}}={{\left( \frac{4}{5} \right)}^{3\times \frac{7}{3}}}={{\left( \frac{4}{5} \right)}^{7}}$

The value of   $\frac{{{2}^{x+3}}\times {{3}^{2x-y}}\times {{5}^{x+y+3}}\times {{6}^{y+1}}}{{{6}^{x+1}}\times {{10}^{y+3}}\times {{15}^{x}}}$ is:

(a) 1

(b) 0

(c) $-1$

(d) 10

(e) None of these

Explanation

$\frac{{{2}^{x+3}}\times {{3}^{2x-y}}\times {{5}^{x+y+3}}\times {{6}^{y+1}}}{{{6}^{x+1}}\times {{10}^{y+3}}\times {{15}^{x}}}=\frac{{{2}^{x=3}}\times {{3}^{2x-y}}\times {{5}^{x+y+3}}\times {{\left( 2\times 3 \right)}^{y+1}}}{{{\left( 2\times 3 \right)}^{x+1}}\times {{\left( 5\times 2 \right)}^{y+3}}\times {{\left( 5\times 3 \right)}^{x}}}$

=$\frac{{{2}^{x+3}}\times {{3}^{2x-y}}\times {{5}^{x+y+3}}\times {{3}^{y+1}}}{{{2}^{x+1}}\times {{3}^{x+1}}\times {{5}^{y+3}}\times {{2}^{y+3}}\times {{3}^{x}}\times {{5}^{x}}}=\frac{{{2}^{x+y+4}}\times {{3}^{2x+1}}\times {{5}^{x+y+3}}}{{{2}^{x+y+4}}\times {{3}^{2x+1}}\times {{5}^{x+y+3}}}=1$

If    ${{\left( \frac{8}{3} \right)}^{2x+1}}\times {{\left( \frac{8}{3} \right)}^{5}}={{\left( \frac{8}{3} \right)}^{x+2}},$ then find the value of $x.$

(a) 0

(b) 2

(c)$-4$

(d) 6

(e) None of these

If ${{9}^{p-1}}+18=3{{h}^{2p-1}}$ then which one of the following options is incorrect?

(a) P is a natural number

(b) P= 2

(c) 1 < p < 4

(d) Can't determine for natural numbers

(e) None of these

Find the value of $\frac{5x{{3}^{n+1}}-4\times {{3}^{n}}}{5\times {{3}^{n+2}}-4\times {{3}^{n+1}}}.$

(a) A natural number which is less than 10

(b) A whole number which is greater than 1 (c) An integer

(d) A fraction whose value is $\frac{1}{3}$

(e) None of these

• ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$
• $\frac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$
• ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}={{\left( {{a}^{n}} \right)}^{m}}$
• If $x$ is a rational number (x > 0) a, b are rational exponents so that a > b then ${{x}^{a}}\div {{x}^{b}}={{x}^{a-b}}.$
• If $x$ is a rational number  $\left( x>0 \right)$ a, b are rational exponents so that a < b then ${{x}^{a}}\div {{x}^{b}}=\frac{1}{{{x}^{b-a}}}$

• "1.23 e 4" means 1.23 times to the fourth power of 10.
• Similarly "5.67 e - 8" means 5.67 divided by eighth power of 10.
• ${{a}^{n}}\to 0asn\to \infty \,when\,\left| a \right|<1$
• ${{a}^{n}}\to 0\,forall\,n,\,if\,a=1$
• ${{0}^{o}}\,is\,in\det er\min ate$
• ${{\left\{ {{\left( {{x}^{a}} \right)}^{b}} \right\}}^{c}}={{x}^{abc}}$

#### Other Topics

LIMITED OFFER HURRY UP! OFFER AVAILABLE ON ALL MATERIAL TILL TODAY ONLY!

You need to login to perform this action.
You will be redirected in 3 sec