Question

1. Find: - (i) \[{{\left( 64 \right)}^{\dfrac{1}{2}}}\] (ii) \[{{32}^{\dfrac{1}{5}}}\] (iii) \[{{125}^{\dfrac{1}{3}}}\]
2. Find: - (i) \[{{9}^{\dfrac{3}{2}}}\] (ii) \[{{32}^{\dfrac{2}{5}}}\] (iii) \[{{16}^{\dfrac{3}{4}}}\]
3. Simplify: - (i) \[{{2}^{\dfrac{2}{3}}}\times {{2}^{\dfrac{1}{5}}}\] (ii) \[{{\left( \dfrac{1}{{{3}^{3}}} \right)}^{7}}\] (iii) \[\dfrac{{{11}^{\dfrac{1}{2}}}}{{{11}^{\dfrac{1}{4}}}}\]

Answer 1

Hint: For part (1) and (2) of the given question, first break the bases of the expressions into prime factors and then apply the formula \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\] to simplify them. For part (3), apply the formulas: - \[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}},\dfrac{1}{{{a}^{m}}}={{a}^{-m}}\] and \[{{a}^{m}}\div {{a}^{n}}={{a}^{m-n}}\] to solve the three questions.

Complete step by step answer:
Here, we have been provided with several questions of the topic ‘exponents and powers’ and we have to simplify them. So, let us check each part one – by – one.
(1). (i) \[{{\left( 64 \right)}^{\dfrac{1}{2}}}\]
Now, 64 can be written as: - 64 = 2 \[\times \] 2 \[\times \]2 \[\times \] 2 \[\times \] 2 \[\times \] 2. Here, 2 is multiplied 6 times, so we can write \[64={{2}^{6}}\]. Therefore, we have,
\[\Rightarrow {{\left( 64 \right)}^{\dfrac{1}{2}}}={{\left( {{2}^{6}} \right)}^{\dfrac{1}{2}}}\]
Applying the identity: - \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\], we get,
\[\begin{align}
  & \Rightarrow {{\left( 64 \right)}^{\dfrac{1}{2}}}={{2}^{6\times \dfrac{1}{2}}} \\
 & \Rightarrow {{\left( 64 \right)}^{\dfrac{1}{2}}}={{2}^{3}} \\
 & \Rightarrow {{\left( 64 \right)}^{\dfrac{1}{2}}}=2\times 2\times 2=8 \\
\end{align}\]
(ii) \[{{32}^{\dfrac{1}{5}}}\]
Now, 32 can be written as: - 32 = 2 \[\times \] 2 \[\times \] 2 \[\times \] 2 \[\times \]2. Here, 2 is multiplied 5 times, so we can write \[32={{2}^{5}}\]. Therefore, we have,
\[\Rightarrow {{\left( 32 \right)}^{\dfrac{1}{5}}}={{\left( {{2}^{5}} \right)}^{\dfrac{1}{5}}}\]
\[\begin{align}
  & \Rightarrow {{\left( 32 \right)}^{\dfrac{1}{5}}}={{2}^{5\times \dfrac{1}{5}}} \\
 & \Rightarrow {{\left( 32 \right)}^{\dfrac{1}{5}}}={{2}^{1}}=2 \\
\end{align}\]
(iii) \[{{\left( 125 \right)}^{\dfrac{1}{3}}}\]
Now, 125 can be written as: - 125 = 5 \[\times \] 5 \[\times \] 5. Here, 5 is multiplied 3 times, so we can write \[125={{5}^{3}}\]. Therefore, we have,
\[\begin{align}
  & \Rightarrow {{\left( 125 \right)}^{\dfrac{1}{3}}}={{\left( {{5}^{3}} \right)}^{\dfrac{1}{3}}} \\
 & \Rightarrow {{\left( 125 \right)}^{\dfrac{1}{3}}}={{5}^{3\times \dfrac{1}{3}}} \\
 & \Rightarrow {{\left( 125 \right)}^{\dfrac{1}{3}}}={{5}^{1}}=5 \\
\end{align}\]
(2) (i) \[{{9}^{\dfrac{3}{2}}}\]
Now, 9 can be written as: - 9 = 3 \[\times \] 3. Here, 3 is multiplied 2 times, so we can write \[9={{3}^{2}}\]. Therefore, we have,
\[\begin{align}
  & \Rightarrow {{9}^{\dfrac{3}{2}}}={{\left( {{3}^{2}} \right)}^{\dfrac{3}{2}}} \\
 & \Rightarrow {{9}^{\dfrac{3}{2}}}={{3}^{2\times \dfrac{3}{2}}} \\
 & \Rightarrow {{9}^{\dfrac{3}{2}}}={{3}^{3}} \\
 & \Rightarrow {{9}^{\dfrac{3}{2}}}=3\times 3\times 3=27 \\
\end{align}\]
(ii) \[{{32}^{\dfrac{2}{5}}}\]
Now, from part (1) (ii) of the given solution above, we get,
\[\begin{align}
  & \Rightarrow {{32}^{\dfrac{2}{5}}}={{\left( {{2}^{5}} \right)}^{\dfrac{2}{5}}} \\
 & \Rightarrow {{32}^{\dfrac{2}{5}}}={{2}^{5\times \dfrac{2}{5}}} \\
 & \Rightarrow {{32}^{\dfrac{2}{5}}}={{2}^{2}} \\
 & \Rightarrow {{32}^{\dfrac{2}{5}}}=2\times 2=4 \\
\end{align}\]
(iii) \[{{16}^{\dfrac{3}{4}}}\]
Now, 16 can be written as: - 16 = 2 \[\times \] 2 \[\times \] 2 \[\times \] 2. Here, 2 is multiplied 4 times, so we can write \[16={{2}^{4}}\]. Therefore, we have,
\[\begin{align}
  & \Rightarrow {{\left( 16 \right)}^{\dfrac{3}{4}}}={{\left( {{2}^{4}} \right)}^{\dfrac{3}{4}}} \\
 & \Rightarrow {{\left( 16 \right)}^{\dfrac{3}{4}}}={{2}^{4\times \dfrac{3}{4}}} \\
 & \Rightarrow {{\left( 16 \right)}^{\dfrac{3}{4}}}={{2}^{3}} \\
 & \Rightarrow {{\left( 16 \right)}^{\dfrac{3}{4}}}=2\times 2\times 2=8 \\
\end{align}\]
(3). (i) \[{{2}^{\dfrac{2}{3}}}\times {{2}^{\dfrac{1}{5}}}\]
Here, we can see that the bases of the two terms multiplied above are the same, so applying the formula: - \[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}\], we get,
\[\begin{align}
  & \Rightarrow {{2}^{\dfrac{2}{3}}}\times {{2}^{\dfrac{1}{5}}}={{2}^{\dfrac{2}{3}+\dfrac{1}{5}}} \\
 & \Rightarrow {{2}^{\dfrac{2}{3}}}\times {{2}^{\dfrac{1}{5}}}={{2}^{\dfrac{\left( 10+3 \right)}{15}}} \\
 & \Rightarrow {{2}^{\dfrac{2}{3}}}\times {{2}^{\dfrac{1}{5}}}={{2}^{\dfrac{13}{15}}} \\
\end{align}\]
(ii) \[{{\left( \dfrac{1}{{{3}^{3}}} \right)}^{7}}\]
Here, applying the formula: - \[\dfrac{1}{{{a}^{m}}}={{a}^{-m}}\], we get,
\[\Rightarrow {{\left( \dfrac{1}{{{3}^{3}}} \right)}^{7}}={{\left( {{3}^{-3}} \right)}^{7}}\]
Now, using the identity: - \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\]
\[\begin{align}
  & \Rightarrow {{\left( \dfrac{1}{{{3}^{3}}} \right)}^{7}}={{3}^{\left( -3 \right)\times 7}} \\
 & \Rightarrow {{\left( \dfrac{1}{{{3}^{3}}} \right)}^{7}}={{3}^{-21}}=\left( \dfrac{1}{{{3}^{21}}} \right) \\
\end{align}\]
(iii) \[\dfrac{{{11}^{\dfrac{1}{2}}}}{{{11}^{\dfrac{1}{4}}}}\]
Here, we can see that the bases of the two terms divided above are the same, so applying the formula: - \[\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}\], we get,
\[\begin{align}
  & \Rightarrow \left( \dfrac{{{11}^{\dfrac{1}{2}}}}{{{11}^{\dfrac{1}{4}}}} \right)={{11}^{\dfrac{1}{2}-\dfrac{1}{4}}} \\
 & \Rightarrow \left( \dfrac{{{11}^{\dfrac{1}{2}}}}{{{11}^{\dfrac{1}{4}}}} \right)={{11}^{\dfrac{\left( 2-1 \right)}{4}}} \\
 & \Rightarrow \left( \dfrac{{{11}^{\dfrac{1}{2}}}}{{{11}^{\dfrac{1}{4}}}} \right)={{11}^{\dfrac{1}{4}}} \\
\end{align}\]

Note:
 One must remember all the basic formulas of the topic ‘exponents and powers’ like: - \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\], \[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}\] and \[\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}\] to solve the above questions. You may note that it is necessary to break the bases of the given expressions into their product of primes. This is done to cancel the fractional powers wherever possible. We can also solve all the questions using logarithm but it is used in higher classes and for some difficult calculations, so in the above questions it is advisable to use the formulas of the topic ‘exponents and powers’.