Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 12 - Exponential Functions and Logarithmic Functions - 12.5 Common Logarithms and Natural Logarithms - 12.5 Exercise Set - Page 819: 87

Answer

$1.442$

Work Step by Step

Using change-of-base, $\displaystyle \log_{b}M=\frac{\log_{a}M}{\log_{a}b}$, we change the base to 10 $\displaystyle \log_{12}36=\frac{\log 36}{\log 12}$ ... next, we recognize $36=6^{2}=2^{2}\cdot 3^{2}$ ... so $\log 36 = 2\log 2+2\log 3$ .... also, $12=3\cdot 4=2^{2}\cdot 3$ ... so $\log 12=2\log 2+\log 3$ ... we applied $\log_{a}M^{p}=p\cdot\log_{a}M$ and $\log_{a}(MN)=\log_{a}M+\log_{a}N$ $... =\displaystyle \frac{2\log 2+2\log 3}{2\log 2+\log 3}$ ... substitute $\log 2=0.301$ and $\log 3=0.477,$ $=\displaystyle \frac{2\times 0.301+2\times 0.477}{2\times 0.301+ 0.477}\approx$1.442075996 To three decimal places, $=1.442$
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.