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  

 

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)  

 

 

  • \[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}\]
  • \[\frac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}\]
  • \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}={{\left( {{a}^{n}} \right)}^{m}}\]
  • 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.
  • \[{{a}^{n}}\to 0asn\to \infty \,when\,\left| a \right|<1\]
  • \[{{a}^{n}}\to 0\,forall\,n,\,if\,a=1\]
  • \[{{0}^{o}}\,is\,in\det er\min ate\]
  • \[{{\left\{ {{\left( {{x}^{a}} \right)}^{b}} \right\}}^{c}}={{x}^{abc}}\]  

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