Answer
$\sqrt[4]{x}\cdot\sqrt[4]{x}=\sqrt{x}$ for $x\ge0$
Work Step by Step
Using $\sqrt[n]{a}\cdot\sqrt[n]{b}=\sqrt[n]{ab}$, the expression $
\sqrt[4]{x}\cdot\sqrt[4]{x}
$ is equivalent to
\begin{align*}
&
\sqrt[4]{x\cdot x}
\\&=
\sqrt[4]{x^2}
\\&=
\sqrt{\sqrt{x^2}}
\\&=
\sqrt{|x|}
&(\text{use }\sqrt{x^2}=|x|)
\\&=
\sqrt{x}
&(\text{since }x\ge0)
.\end{align*}
Hence, $\sqrt[4]{x}\cdot\sqrt[4]{x}=\sqrt{x}$ for $x\ge0$.
