Negative Rational Number as Exponent
Category : 7th Class
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
Answer: (a)
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
Answer: (b)
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
Answer: (a)
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
Answer: (a)
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
Answer: (d)
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
Answer: (a)
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
Answer: (b)
Evaluate: \[{{\left[ {{\left( 729 \right)}^{\frac{-5}{3}}} \right]}^{-\frac{1}{2}}}\]
(a) 243
(b) 81
(c) 27
(d) 9
(e) None of these
Answer: (a)
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
Answer: (b)
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
Answer: (c)
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
Answer: (c)
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
Answer: (a)
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
Answer: (c)
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
Answer: (d)
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
Answer: (d)
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