## Calculus with Applications (10th Edition)

For $a > 0, a\neq 1$, and $x > 0,\ \qquad y=\log_{a}x$ means $a^{y}=x$. So, $a^{\log_{a}x}=x$ Also, for $a=e,\ \log_{a}x$ is written as $\ln x.$ Therefore, $e^{\ln 2}=2.$ The statement is true.