We have seen that raising to a power is the inverse operation for extracting roots, so that
\begin{equation*}
(\sqrt[n]{a})^n = a
\end{equation*}
For example,
\begin{equation*}
(\sqrt[4]{16})^4 = 16~~~~~~\text{and}~~~~~~(\sqrt[5]{x})^5 = x
\end{equation*}
What if the power and root operations occur in the opposite order? Is it always true that \(~\sqrt[n]{a^n} = a~\text{?}\)
First, consider the case where the index is an odd number. For example,
\begin{equation*}
\sqrt[3]{2^3} = \sqrt[3]{8} = 2 ~~~~~~\text{and}~~~~~~\sqrt[3]{(-2)^3} = \sqrt[3]{-8} = -2
\end{equation*}
Because every real number has exactly one \(n\)th root if \(n\) is odd, we see that,
\begin{equation*}
\sqrt[n]{a^n} = a,~~~~\text{for}~n~\text{odd}
\end{equation*}
However, if \(n\) is even, then \(a^n\) is positive, regardless of whether \(a\) itself is positive or negative, and hence \(\sqrt[n]{a^n}\) is positive also. For example, if \(a=-3\text{,}\) then
\begin{equation*}
\sqrt{(-3)^2} = \sqrt{9} = 3
\end{equation*}
In this case, \(\sqrt{a^2}\) does not equal \(a\text{,}\) because \(a\) is negative but \(\sqrt{a^2}\) is positive. We must be careful when taking even roots of powers. We have the following special relationship for even roots.
\begin{equation*}
\blert{\sqrt[n]{a^n} = \abs{a},~~~~~\text{for}~n~\text{even}}
\end{equation*}
We summarize our results in the box below.
