## Precalculus (10th Edition)

$y=Cx(x+1)$
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 $y=Cx(x+1)$