Answer
\[x=\ln \left( \frac{{{e}^{2}}-1}{2} \right)\]
Work Step by Step
\[\begin{align}
& f\left( x \right)={{e}^{x}}\text{ on }\left[ 0,2 \right] \\
& \text{The average is given by} \\
& {{f}_{avg}}=\frac{1}{b-a}\int_{a}^{b}{f\left( x \right)dx} \\
& \text{Therefore,} \\
& {{f}_{avg}}=\frac{1}{2-0}\int_{0}^{2}{{{e}^{x}}dx} \\
& {{f}_{avg}}=\frac{1}{2}\int_{0}^{2}{{{e}^{x}}dx} \\
& \text{Integrating} \\
& {{f}_{avg}}=\frac{1}{2}\left[ {{e}^{x}} \right]_{0}^{2} \\
& {{f}_{avg}}=\frac{1}{2}\left[ {{e}^{2}}-{{e}^{0}} \right] \\
& {{f}_{avg}}=\frac{{{e}^{2}}-1}{2} \\
& \text{The point at which the given function equals its average value is} \\
& {{e}^{x}}=\frac{{{e}^{2}}-1}{2} \\
& \ln {{e}^{x}}=\ln \left( \frac{{{e}^{2}}-1}{2} \right) \\
& x=\ln \left( \frac{{{e}^{2}}-1}{2} \right) \\
\end{align}\]
