Answer
$\frac{1}{3}\ln(x^{4}+12)-\ln(x+16)-\frac{1}{2}\ln(x-3)$
Work Step by Step
We expand:
$\ln(\frac{\sqrt[3]{x^{4}+12}}{(x+16)\sqrt{x-3}})$
$\ln(x^{4}+12)^{\frac{1}{3}}-[\ln(x+16)+\ln(x-3)^{\frac{1}{2}}]$
$\frac{1}{3}\ln(x^{4}+12)-[\ln(x+16)+\frac{1}{2}\ln(x-3)]$
$\frac{1}{3}\ln(x^{4}+12)-\ln(x+16)-\frac{1}{2}\ln(x-3)$
