Question

How do you write the expression \[{\left( 4 \right)^{\dfrac{4}{3}}}\] in radical form?

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.

Complete step by step answer:
In the question, the expression to be solved is \[{\left( 4 \right)^{\dfrac{4}{3}}}\].
Now, we find the prime factorization of \[4\],
\[\begin{array}{l}2\left| \!{\underline {\,
  4 \,}} \right. \\2\left| \!{\underline {\,
  2 \,}} \right. \\{\rm{ }}\left| \!{\underline {\,
  1 \,}} \right. \end{array}\]
Hence, \[4 = 2 \times 2 = {2^2}\]
\[{\left( 4 \right)^{\dfrac{4}{3}}} = {\left( {{2^2}} \right)^{\dfrac{4}{3}}}\]
Now, we know, \[{\left( {{a^m}} \right)^n} = {a^{m \times n}}\]
\[{\left( {{2^2}} \right)^{\dfrac{4}{3}}} = {2^{2 \times \dfrac{4}{3}}} = {2^{\dfrac{8}{3}}}\]
Now,
\[{2^8} = 256\]
So, we have,
\[{2^{\dfrac{8}{3}}} = {\left( {256} \right)^{\dfrac{1}{3}}}\]
Also, we know that,
\[256 = 64 \times 4 = {4^3} \times 4\]
Thus,
\[{\left( {256} \right)^{\dfrac{1}{3}}} = {\left( {{4^3} \times 4} \right)^{\dfrac{1}{3}}} = 4\sqrt[3]{4}\]

Hence, the answer is \[4\sqrt[3]{4}\].

Note:
We might get the answer wrong if we do not know the correct meaning of the negative power. A negative power does not affect the sign of the base. 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.