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 - Page 359: 38


$\log\dfrac{y^{3}}{\sqrt{2x}}=3\log y-\dfrac{1}{2}\log 2-\dfrac{1}{2}\log x$

Work Step by Step

$\log\dfrac{y^{3}}{\sqrt{2x}}$ We begin expanding, by noting that a division can be expanded as a substraction: $\log\dfrac{y^{3}}{\sqrt{2x}}=\log y^{3}-\log\sqrt{2x}=...$ Rewrite the expression obtained like this: $...=\log y^{3}-\log(2x)^{1/2}=...$ The exponents present in $\log y^{3}$ and $\log(2x)^{1/2}$ can be taken to the front of their respective logarithms to multiply: $...=3\log y-\dfrac{1}{2}\log(2x)=...$ Finally, the logarithm of a product can be expanded as a sum: $...=3\log y-\dfrac{1}{2}(\log 2+\log x)=...$ $...=3\log y-\dfrac{1}{2}\log 2-\dfrac{1}{2}\log x$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.