## Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole

# Chapter 4 - Review - Exercises: 50

#### 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: $...=\dfrac{1}{3}\ln(x^{4}+12)-\ln(x+16)-\dfrac{1}{2}\ln(x-3)$

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