## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$4$
$\because \log_a M = \dfrac{\log_b M}{\log_b a}$ $\therefore \log_{e^2} 16 = \dfrac{\log_e 16}{\log_e e^2}$ Recall that $\log_e a = \ln a$ Thus, $\dfrac{\log_e 16}{\log_e e^2} =\dfrac{\ln{ 16}}{\ln{e^2}}$ So the given expression si equivalent to: $e^{\log_{e^2} 16} = e^{\left(\frac{\ln{ 16}}{\ln{e^2}}\right)}$ Since $\ln e^x = x$, then the expression above simplifies to: $e^{\left(\frac{\ln{ 16}}{\ln{e^2}}\right)} = e^{\left(\frac{\ln{16}}{2} \right)}= e^{0.5 \ln{16}}$ With $e^{x \ln a}=a^x$, then $e^{0.5 \ln{16}} =16^{0.5} = \sqrt{16}=\boxed{4}$