## Precalculus (10th Edition)

$\frac{1}{2}$
The definition of the logarithmic function says that $y=\log_a{x}$ if and only if $a^y=x$. Also, $a\gt0,a\ne1$ and $x\gt0$. Hence $\ln{\sqrt{e}}=\log_{e} {\sqrt{e}}=y$, then $\left(e\right)^y=\sqrt{e}$ and we know that $\sqrt{e}=e^{\frac{1}{2}}.$ Thus, $\left(e\right)^{y}=\left(e\right)^{\frac{1}{2}}$. We know that $a^b=a^c\longrightarrow b=c$ if $a\ne1,a\ne-1$ (which applies here), hence $y=\frac{1}{2}$.