Answer
$\ln{\left(\frac{8x^2}{(x+4)^{\frac{1}{2}}}\right)}$
Work Step by Step
RECALL:
(1) $\ln{P} + \ln{Q} = \ln{(PQ)}$
(2) $\ln{P} - \ln{Q} = \ln{(\frac{P}{Q})}$
(3) $a(\ln{x}) = \ln{(x^a)}$
Use rule (3) above to obtain:
$=\ln{(2^3)}+\ln{(x^2)}-\ln{[(x+4)^{\frac{1}{2}}]}
\\=\ln{8}+\ln{(x^2)}-\ln{[(x+4)^{\frac{1}{2}}]}$
Use rule (1) above to obtain:
$=\ln{(8x^2)}-\ln{[(x+4)^{\frac{1}{2}}]}$
Use rule (2) above to obtain:
$=\ln{\left(\frac{8x^2}{(x+4)^{\frac{1}{2}}}\right)}$