Answer
$y=Cx(x+1)$
Work Step by Step
We know that $\log_a {x}+\log_a {y}=\log_a {(xy)}$,
Hence,
$\ln{x}+\ln({x+1})+\ln{C}\\
=\ln{[x(x+1)(C)]}\\
=\ln{[Cx(x+1)]}$
Thus, the given equation is equivalent to
$\ln{y} = \ln{[Cx(x+1)]}$
Use the rule $\log_a{M}=\log_a{N} \longrightarrow M=N$ to obtain
$y=Cx(x+1)$
