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This is the second of our many quizzes on evaluating exponents. Click on this to access more such competitive quizzes on exponents.

Evaluating Exponents Quiz 2

mathematics

 Simplify the expression ${\left( {\dfrac{{256}}{{81}}} \right)^{ - \frac{3}{4}}} \times \left[ {{{\left( {\dfrac{{100}}{{49}}} \right)}^{ - \frac{3}{2}}} \div {{\left( {\dfrac{{10}}{6}} \right)}^{ - 3}}} \right]$ and determine the equivalent value.

 ${({a^m})^n} = {a^{mn}}$ use this identity to simplify the given expression.

  $\text{According to the question; }\\{\left( {\dfrac{{256}}{{81}}} \right)^{ - \frac{3}{4}}} \times \left[ {{{\left( {\dfrac{{100}}{{49}}} \right)}^{ - \frac{3}{2}}} \div {{\left( {\dfrac{{10}}{6}} \right)}^{ - 3}}} \right] \\ \Rightarrow { {\left( {\dfrac{{81}}{{256}}} \right)} ^{\frac{3}{4}}} \times \left[ {\dfrac{{{{\left( {\dfrac{{49}}{{100}}} \right)}^{\frac{3}{2}}}}}{{{{\left( {\dfrac{6}{{10}}} \right)}^3}}}} \right] \\ \Rightarrow {\left( {\dfrac{3}{4}} \right)^4} \times \left[ {\dfrac{{{{\left( {\dfrac{7}{{10}}} \right)}^{2 \times \frac{3}{2}}}}}{{{{\left( {\dfrac{6}{{10}}} \right)}^3}}}} \right] \\ \Rightarrow {\left( {\dfrac{3}{4}} \right)^3} \times \left[ {{{\left( {\dfrac{7}{6}} \right)}^3}} \right] \\ \Rightarrow \left( {\dfrac{{27}}{{64}} \times \dfrac{{343}}{{216}}} \right) \Rightarrow \dfrac{{343}}{{512}} $ 

 When we simplify the expression: $\dfrac{{{4^{t + 3}} - 4({4^t})}}{{{4^{(4t + 4)}}}}$,where $t = 1$ value will be equal to:

 Simplify the expression given in the question by putting the value of $t = 1$.

  $\text{As, }\\ \dfrac{{{4^{t + 3}} - 4({4^t})}}{{{4^{(4t + 4)}}}}\\ \text{ Putting  }t = 1,\text{ in above expression}\\ \dfrac{{{4^{1 + 3}} - 4({4^1})}}{{{4^{(4(1) + 4)}}}} \\ \Rightarrow \dfrac{{{4^4} - {4^2}}}{{{4^8}}} \\ \Rightarrow \dfrac{{{4^2}(15)}}{{{4^8}}} \\ \Rightarrow {4^{ - 6}}(15) $ 

 What will be the value of ${\left( {\dfrac{{\sqrt 2 }}{5}} \right)^8} \div {\left( {\dfrac{{\sqrt 2 }}{5}} \right)^{13}}$

 $\dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b} \times \dfrac{d}{c}$ use this property to simplify the above expression.

 ${\left( {\dfrac{{\sqrt 2 }}{5}} \right)^8} \div {\left( {\dfrac{{\sqrt 2 }}{5}} \right)^{13}} \\ \Rightarrow \dfrac{{{{(\sqrt 2 )}^8}}}{{{5^8}}} \times \dfrac{{{5^{13}}}}{{{{(\sqrt 2 )}^{13}}}} \\ \Rightarrow \dfrac{{{5^{13 - 8}}}}{{{{(\sqrt 2 )}^{13 - 8}}}} \\ \Rightarrow {\left( {\dfrac{5}{{\sqrt 2 }}} \right)^5} $ 

 Arrange the $a,b,c,d$ in the increasing order of magnitude, here $a,b,c,d$ are  $\sqrt[3]{4},\sqrt[6]{{10}},\sqrt[{12}]{{50}},\sqrt[6]{{15}}$.

 Take the LCM of the order of square roots and simplify.

 $\sqrt[3]{4},\sqrt[6]{{10}},\sqrt[{12}]{{50}},\sqrt[6]{{15}}\\ \text{LCM of 3,6,12 is 12 So, }\\ \sqrt[{12}]{{{4^3}}},\sqrt[{12}]{{{{10}^2}}},\sqrt[{12}]{{50}},\sqrt[{12}]{{{{15}^2}}} \\  \Rightarrow \sqrt[{12}]{{64}},\sqrt[{12}]{{100}},\sqrt[{12}]{{50}},\sqrt[{12}]{{225}} \\ \text{Now we can compare the above values, }\\ \text{Hence }d > b > a > c$

 If value of $pqr = 1$, then evaluate the following expression $\dfrac{1}{{1 + p + {q^{ - 1}}}} + \dfrac{1}{{1 + q + {r^{ - 1}}}} + \dfrac{1}{{1 + r + {p^{ - 1}}}}$, is equal to 

 As $pqr = 1$, we have to use $r = \dfrac{1}{{pq}}$ to make similar terms.

 $\text{Put, }r = \dfrac{1}{{pq}} \\ \Rightarrow \dfrac{1}{{(1 + p + \dfrac{1}{q})}} + \dfrac{1}{{(1 + q + \dfrac{1}{r})}} + \dfrac{1}{{(1 + r + \dfrac{1}{p})}} \\ \Rightarrow \dfrac{1}{{(1 + p + \dfrac{1}{q})}} + \dfrac{1}{{(1 + q + pq)}} + \dfrac{1}{{(1 + \dfrac{1}{{pq}} + \dfrac{1}{p})}} \\ \Rightarrow \dfrac{q}{{(q + pq + 1)}} + \dfrac{1}{{(1 + q + pq)}} + \dfrac{{pq}}{{(1 + q + pq)}} \\ \Rightarrow \dfrac{{(q + 1 + pq)}}{{(1 + q + pq)}} \Rightarrow 1$