Answer
$4$
Work Step by Step
$\because \log_a M = \dfrac{\log_b M}{\log_b a}$
$\therefore \log_{e^2} 16 = \dfrac{\log_e 16}{\log_e e^2}$
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
$e^{0.5 \ln{16}} =16^{0.5} = \sqrt{16}=\boxed{4}$
