Answer
$3$
Work Step by Step
$\because \log_a M = \dfrac{\log_b M}{\log_b a}$
$\therefore \log_{e^2} 9 = \dfrac{\log_e 9}{\log_e e^2}$
Note that:
$\log_e a = \ln a$
Thus,
$\dfrac{\log_e 9}{\log_e e^2} =\dfrac{\ln{ 9}}{\ln{e^2}}$
So the given expression is equivalent to:
$e^{\log_{e^2} 16} = e^{\left(\frac{\ln{ 9}}{\ln{e^2}}\right)}$
Recall that $\ln e^x = x$
Hence,
$\ln{e^2} = 2$
Therefore,
$e^{\left(\frac{\ln{ 9}}{\ln{e^2}}\right)} = e^{\left(\frac{\ln{9}}{2}\right)}= e^{0.5 \ln{9}}$
With $e^{x \ln a}=a^x $, then:
$e^{0.5 \ln{9}} =9^{0.5} = \sqrt9=\boxed{3}$
