## Precalculus (10th Edition)

$y=Ce^{5x}-4$.
We know that if $e^x=y$, then $\ln {y}=x$, hence because $5x=\ln{(e^{5x})}$, $\ln{(y+4)}=\ln{(e^{5x})}+\ln{(C)}$. We know that $\log_a {x}+\log_a {y}=\log_a {(x\cdot y)}$, hence $\ln{(y+4)}=\ln{(e^{5x})}+\ln{(C)}=\ln{(Ce^{5x})}.$ 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 $y+4=Ce^{5x}$, thus $y=Ce^{5x}-4$.