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Exponent rules Quiz 2

mathematics

 Solve for $x: \dfrac{{{6}^{x}}\times {{2}^{2x}}\times {{3}^{2x}}}{{{6}^{4}}}=\dfrac{{{3}^{-7}}}{{{2}^{7}}\times 18\times 12}$ 

 Change the right side of the given expression into the power of prime factors of 6 i.e., 2 and 3. Then, solve for $x$ by equating the powers of both the sides.

  $\dfrac{{{6}^{x}}\times {{2}^{2x}}\times {{3}^{2x}}}{{{6}^{4}}}=\dfrac{{{3}^{-7}}}{{{2}^{7}}\times 18\times 12} \\ \Rightarrow {{6}^{3x-4}}=\dfrac{{{3}^{-7}}}{{{2}^{7}}\times {{2}^{3}}\times {{3}^{3}}}\,\,\,\,\,\,\,\left[ \because {{a}^{m}}\times {{a}^{n}}={{a}^{m+n}},{{a}^{m}}\times {{b}^{m}}={{(ab)}^{m}},\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}},{{a}^{-1}}=\dfrac{1}{a} \right] \\ \Rightarrow {{6}^{3x-4}}={{6}^{-10}} \\ \Rightarrow 3x-4=-10\left[ \because m=n\text{ when }{{a}^{m}}={{a}^{n}} \right] \\ \Rightarrow x=-2$

 Simplify: $\dfrac{{{4}^{3n}}\times {{16}^{2(3n+1)}}}{{{4}^{5n-1}}\times {{16}^{3}}\times {{16}^{5n}}}$

 Use exponent rule $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$ and then convert 16 into the power of 4. Then, use rule ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ to simplify. 

  $\dfrac{{{4}^{3n}}\times {{16}^{2(3n+1)}}}{{{4}^{5n-1}}\times {{16}^{3}}\times {{16}^{5n}}} \\ \Rightarrow {{4}^{3n-5n+1}}\times {{16}^{6n+2-3-5n}}\,\,\,\,\,\,\,\left[ \because \dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}},{{a}^{-1}}=\dfrac{1}{a} \right] \\ \Rightarrow {{4}^{-2n+1}}\times {{\left( {{\text{4}}^{2}} \right)}^{n-1}}\text{                                     } \\ \Rightarrow {{4}^{-2n+1+2n-2}} \\ \Rightarrow \dfrac{1}{4}\,\,\,\,\,\,\,\left[ \because {{a}^{m}}\times {{a}^{n}}={{a}^{m+n}} \right]$ 

 Simplify: $\dfrac{{{p}^{{{({{a}^{2}}+{{b}^{2}})}^{2}}}}}{{{p}^{2{{a}^{2}}{{b}^{2}}}}}\times \dfrac{1}{{{\left( {{p}^{({{a}^{2}}+{{c}^{2}})}} \right)}^{({{a}^{2}}-{{c}^{2}})}}}\times \dfrac{{{p}^{2{{b}^{2}}{{c}^{2}}}}}{{{p}^{{{({{b}^{2}}+{{c}^{2}})}^{2}}}}}$

 First use rule $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$ and then use formula ${{a}^{4}}+{{b}^{4}}={{({{a}^{2}}+{{b}^{2}})}^{2}}-2{{a}^{2}}{{b}^{2}}$. Then, add all the terms in the exponent to get the answer.

  $\dfrac{{{p}^{{{({{a}^{2}}+{{b}^{2}})}^{2}}}}}{{{p}^{2{{a}^{2}}{{b}^{2}}}}}\times \dfrac{1}{{{\left( {{p}^{({{a}^{2}}+{{c}^{2}})}} \right)}^{({{a}^{2}}-{{c}^{2}})}}}\times \dfrac{{{p}^{2{{b}^{2}}{{c}^{2}}}}}{{{p}^{{{({{b}^{2}}+{{c}^{2}})}^{2}}}}} \\ \Rightarrow {{p}^{{{({{a}^{2}}+{{b}^{2}})}^{2}}-2{{a}^{2}}{{b}^{2}}}}\times {{p}^{-({{a}^{2}}+{{c}^{2}})({{a}^{2}}-{{c}^{2}})}}\times {{p}^{-({{({{b}^{2}}+{{c}^{2}})}^{2}}-2{{b}^{2}}{{c}^{2}})}}\text{              }\left[ \because \dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}},{{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}} \right] \\ \Rightarrow {{p}^{{{a}^{4}}+{{b}^{4}}}}\times {{p}^{-{{a}^{4}}+{{c}^{4}}}}\times {{p}^{-{{b}^{4}}-{{c}^{4}}}}\text{                                             }\left[ \because {{a}^{4}}+{{b}^{4}}={{({{a}^{2}}+{{b}^{2}})}^{2}}-2{{a}^{2}}{{b}^{2}} \right]\,\,\,\,\,\,\, \\ \Rightarrow {{p}^{{{a}^{4}}+{{b}^{4}}-{{a}^{4}}+{{c}^{4}}-{{b}^{4}}-{{c}^{4}}}}\Rightarrow {{p}^{0}}\Rightarrow 1\text{                                        }\left[ \because {{a}^{m}}\times {{a}^{n}}={{a}^{m+n}} \right]$ 

 If $s=3\text{ and }t=27$, then find the value of $\dfrac{1}{({{s}^{n}}+{{t}^{n}})({{s}^{n}}-{{t}^{n}})}\left( {{\left( {{s}^{n}} \right)}^{2}}\sqrt{\dfrac{{{t}^{2}}}{{{s}^{2}}}}-\dfrac{{{t}^{2n+1}}}{s} \right)$ 

 First simplify the expression by converting ${{\left( {{s}^{n}} \right)}^{2}}={{s}^{2n}}$ and taking $t$ common from the brackets. Then, use formula ${{a}^{2}}-{{b}^{2}}=(a+b)(a-b)$. Then, put values of $s$ and $t$. 

  $\dfrac{1}{({{s}^{n}}+{{t}^{n}})({{s}^{n}}-{{t}^{n}})}\left( {{\left( {{s}^{n}} \right)}^{2}}\sqrt{\dfrac{{{t}^{2}}}{{{s}^{2}}}}-\dfrac{{{t}^{2n+1}}}{s} \right) \\ \Rightarrow \dfrac{1}{({{s}^{2n}}-{{t}^{2n}})}\left( \dfrac{{{s}^{2n}}t-{{t}^{2n}}t}{s} \right)\text{     }\left[ \because {{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}} \right]\text{     } \\ \Rightarrow \dfrac{t}{s}\Rightarrow \dfrac{27}{3}\Rightarrow 9$  

 If ${{9}^{\sqrt{y}}}=360\text{ and }{{7}^{4x}}=60$, then find the value of $\dfrac{{{27}^{1-\sqrt{4y}}}}{{{\left( {{7}^{-2x}} \right)}^{8}}}$.

 First convert the given expression in ${{9}^{\sqrt{y}}}$ and ${{7}^{4x}}$, then put the given values. Then, solve using exponent rules.  

  $\dfrac{{{27}^{1-\sqrt{4y}}}}{{{\left( {{7}^{-2x}} \right)}^{8}}} \\ \Rightarrow \dfrac{27\times {{\left( {{3}^{3}} \right)}^{-2\sqrt{y}}}}{{{\left( {{7}^{4x}} \right)}^{-4}}}\text{                                  }\left[ \because {{a}^{m-n}}={{a}^{m}}\times {{a}^{-n}},{{\left( {{a}^{m}} \right)}^{n}}={{\left( {{a}^{n}} \right)}^{m}}={{a}^{mn}} \right] \\ \Rightarrow \dfrac{{{3}^{3}}\times {{\left( {{3}^{2}} \right)}^{-3\sqrt{y}}}}{{{60}^{-4}}}\Rightarrow \dfrac{{{3}^{3}}\times {{\left( {{9}^{\sqrt{y}}} \right)}^{-3}}}{{{60}^{-4}}} \\ \Rightarrow \dfrac{{{3}^{3}}\times {{36}^{-3}}\times {{10}^{-3}}}{{{6}^{-4}}\times {{10}^{-4}}}\Rightarrow \dfrac{{{3}^{3}}\times {{36}^{-3}}\times {{10}^{-3}}}{{{36}^{-2}}\times {{10}^{-4}}} \\ \Rightarrow \dfrac{{{3}^{3}}\times 10}{36}\Rightarrow \dfrac{15}{2}\text{                                       }\left[ \because \dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}} \right] $