Question

How do you find the simplified radical form of \[{\left( {32} \right)^{ - \dfrac{3}{5}}}\]?

Answer 1

Hint: In the given question, we have been asked to calculate a given expression. To solve the question, we need to know how to convert a negative exponential power to a positive exponential power. We do that, and then we just solve the exponent as normal.

Formula Used:
To solve the question, we are going to use the formula to convert a negative exponential power to a positive exponential power, which is,
\[{a^{ - b}} = \dfrac{1}{{{a^b}}}\]

Complete step-by-step answer:
In the question, the expression to be solved is \[{\left( {32} \right)^{ - \dfrac{3}{5}}}\].
First, we convert the negative power to positive,
\[{\left( {32} \right)^{ - \dfrac{3}{5}}} = \dfrac{1}{{{{\left( {32} \right)}^{\dfrac{3}{5}}}}}\]
Now, we find the prime factorization of \[32\],
\[\begin{array}{l}2\left| \!{\underline {\,
  {32} \,}} \right. \\2\left| \!{\underline {\,
  {16} \,}} \right. \\2\left| \!{\underline {\,
  8 \,}} \right. \\2\left| \!{\underline {\,
  4 \,}} \right. \\2\left| \!{\underline {\,
  2 \,}} \right. \\{\rm{ }}\left| \!{\underline {\,
  1 \,}} \right. \end{array}\]
Hence, \[32 = 2 \times 2 \times 2 \times 2 \times 2 = {2^5}\]
\[\dfrac{1}{{{{\left( {32} \right)}^{\dfrac{3}{5}}}}} = \dfrac{1}{{{{\left( {{2^5}} \right)}^{\dfrac{3}{5}}}}}\]
Now, we know, \[{\left( {{a^m}} \right)^n} = {a^{m \times n}}\]
\[\dfrac{1}{{{{\left( {{2^5}} \right)}^{\dfrac{3}{5}}}}} = \dfrac{1}{{\left( {{2^{{5} \times \dfrac{3}{{{5}}}}}} \right)}} = \dfrac{1}{{\left( {{2^3}} \right)}} = \dfrac{1}{8}\]
Hence, the answer is \[\dfrac{1}{8}\]

Note:The negative power only affects the fraction kind of thing of the number. It does not change anything about the sign with the number. So, if we have a negative power, we just take the reciprocal of the number, and calculate the number normally.