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 - 12.4 Properties of Logarithmic Functions - 12.4 Exercise Set - Page 810: 59

Answer

$1.953$

Work Step by Step

Using the properties of logarithms, the given expression, $ \log_b 15 ,$ is equivalent to \begin{array}{l}\require{cancel} \log_b (3\cdot5) \\\\= \log_b 3+\log_b 5 .\end{array} Since it is given that $ \log_b 3=0.792 $ and $ \log_b5=1.161 $, the expression above, $ \log_b 3+\log_b 5 ,$ evaluates to \begin{array}{l}\require{cancel} 0.792+1.161 \\\\= 1.953 .\end{array}
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.