Answer
$\ln\Big(\dfrac{\sqrt[3]{x^{4}+12}}{(x+16)\sqrt{x-3}}\Big)=\dfrac{1}{3}\ln(x^{4}+12)-\ln(x+16)-\dfrac{1}{2}\ln(x-3)$
Work Step by Step
$\ln\Big(\dfrac{\sqrt[3]{x^{4}+12}}{(x+16)\sqrt{x-3}}\Big)$
Expand the logarithm of the division as a subtraction:
$\ln\Big(\dfrac{\sqrt[3]{x^{4}+12}}{(x+16)\sqrt{x-3}}\Big)=\ln\sqrt[3]{x^{4}+12}-\ln(x+16)\sqrt{x-3}=...$
Rewrite the expression by changing the roots to powers with rational exponents:
$...=\ln(x^{4}+12)^{1/3}-\ln(x+16)(x-3)^{1/2}=...$
Expand $\ln(x+16)(x-3)^{1/2}$ as a sum:
$...=\ln(x^{4}+12)^{1/3}-[\ln(x+16)+\ln(x-3)^{1/2}]=...$
Remove the brackets by changing the sign of the terms inside them:
$...=\ln(x^{4}+12)^{1/3}-\ln(x+16)-\ln(x-3)^{1/2}=...$
Take the exponents to multiply in front of their respective logarithm:
$...=\dfrac{1}{3}\ln(x^{4}+12)-\ln(x+16)-\dfrac{1}{2}\ln(x-3)$
