Answer
$\ln{[(x-4)^{\frac{1}{2}}(x^2+4x)^{\frac{5}{2}}]}$
Work Step by Step
Distribute $\frac{1}{2}$ to obtain:
$=\frac{1}{2}\ln{(x-4)}+\frac{5}{2}\ln{(x^2+4x)}$
RECALL:
(1) $n \cdot \ln{P}=\ln{(P^n)}$
(2) $\ln{P} + \ln{Q}=\ln{(PQ)}$
(3) $\ln{P} - \ln{Q}=\ln{(\frac{P}{Q})}$
(4) $a^m \cdot a^n=a^{m+n}$
Use rule (1) above to obtain
$=\ln{(x-4)^{\frac{1}{2}}}+\ln{(x^2+4x)^{\frac{5}{2}}}$
Use rule (2) above to obtain:
$=\ln{[(x-4)^{\frac{1}{2}}(x^2+4x)^{\frac{5}{2}}]}$
