Answer
$\log{x}+log{(x+2)}-2\cdot \log{(x+3)}.$
Work Step by Step
Recall:
(1) $\sqrt[m]{a}=a^{\frac{1}{m}}$
(2) $\log_a {x^n}=n\cdot \log_a {x}$.
(3) $\log_a{xy}=\log_a{x} +\log_a{y}$
(4) $\log_a{\frac{x}{y}}=\log_a{x} -\log_a{y}$
Use Rule (4) to obtain: $\log{\frac{x(x+2)}{(x+3)^2}}=\log{x(x+2)}-\log{(x+3)^2}$.
Use Rule (2) to obtain:
$\log{x(x+2)}-\log{(x+3)^2}=\log{(x(x+2))}-2\cdot \log{(x+3)}.$
Use Rule (3) to obtain:
$\log{(x(x+2))}-2\cdot \log{(x+3)}=\log{x}+log{(x+2)}-2\cdot \log{(x+3)}.$
