Answer
$y=3+Ce^{-4x}$.
Work Step by Step
We know that if $e^x=y$, then $\ln {y}=x$, hence because $-4x=\ln{(e^{-4x})}$, $\ln{(y-3)}=\ln{(e^{-4x})}+\ln{(C)}$.
We know that $\log_a {x}+\log_a {y}=\log_a {(x\cdot y)}$, hence $\ln{(y-3)}=\ln{(e^{-4x})}+\ln{(C)}=\ln{(Ce^{-4x})}.$
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 $y-3=Ce^{-4x}$, thus $y=3+Ce^{-4x}$.
