Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 4 - Section 4.4 - Laws of Logarithms - 4.4 exercises: 47

Answer

$\log\sqrt{\dfrac{x^{2}+4}{(x^{2}+1)(x^{3}-7)^{2}}}=\dfrac{1}{2}[\log(x^{2}+4)-\log(x^{2}+1)-2\log(x^{3}-7)]$

Work Step by Step

$\log\sqrt{\dfrac{x^{2}+4}{(x^{2}+1)(x^{3}-7)^{2}}}$ Before we begin with the expansion process, let's rewrite the expression like this: $\log[\dfrac{x^{2}+4}{(x^{2}+1)(x^{3}-7)^{2}}]^{1/2}$ The exponent in this expression can be taken to the front of the logarithm to multiply: $\log[\dfrac{x^{2}+4}{(x^{2}+1)(x^{3}-7)^{2}}]^{1/2}=\dfrac{1}{2}\log\dfrac{x^{2}+4}{(x^{2}+1)(x^{3}-7)^{2}}=...$ The logarithm of a division can be expanded as a substraction: $...=\dfrac{1}{2}[\log(x^{2}+4)-\log(x^{2}+1)(x^{3}-7)^{2}]=...$ The logarithm of a product can be expanded as a sum: $...=\dfrac{1}{2}[\log(x^{2}+4)-[\log(x^{2}+1)+\log(x^{3}-7)^{2}]]=...$ $...=\dfrac{1}{2}[\log(x^{2}+4)-\log(x^{2}+1)-\log(x^{3}-7)^{2}]=...$ The exponent present in $\log(x^{3}-7)^{2}$ can be taken to the front of its respective logarithm to multiply: $...=\dfrac{1}{2}[\log(x^{2}+4)-\log(x^{2}+1)-2\log(x^{3}-7)]$
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