Answer
$$
e^{-x} =\left(e^{4}\right)^{x+3}
$$
The solution of the given equation is
$$
x=-\frac{12}{5} .
$$
Work Step by Step
$$
e^{-x} =\left(e^{4}\right)^{x+3}
$$
Since the bases must be the same, write 16 as $2^{2 }$ and 64 as $ 2^{6} $ giving
$$
\begin{aligned} e^{-x} &=\left(e^{4}\right)^{x+3} \\ e^{-x} &=e^{4 x+12} \\-x &=4 x+12 \\-5 x &=12 \\ x &=-\frac{12}{5} \end{aligned}
$$
So, the solution of the given equation is
$$
x=-\frac{12}{5} .
$$
