Answer
$y=Ce^{-2x}$.
Work Step by Step
We know that if $e^x=y$, then $\ln {y}=x$, hence because $-2x=\ln{(e^{-2x})}$, $\ln{y}=\ln{e^{-2x}} + \ln{C}$.
We know that $\log_a {x}+\log_a {y}=\log_a {(x\cdot y)}$, hence $\ln{(y)}=\ln{(e^{-2x})}+\ln{(C)}=\ln{(Ce^{-2x})}.$
The base is same on the 2 sides on the equation (and it is not $1$), hence they will be equal if the exponents are equal.($\log_a{M}=\log_a{N} \longrightarrow M=N$.)
Hence $y=Ce^{-2x}$.
