College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 4 - Exponential and Logarithmic Functions - Exercise Set 4.3 - Page 477: 41

Answer

1

Work Step by Step

We know that $log(x)$ is a common logarithm with an understand base of 10. Therefore, $log5+log2=log_{10}5+log_{10}2$. Based on the product rule of logarithms, we know that $log_{b}(MN)=log_{b}M+log_{b}N$ (for $M\gt0$ and $N\gt0$). Therefore, $log_{10}5+log_{10}2=log_{10}(5\times2)=log_{10}10$. Based on the definition of the logarithmic function, we know that $y=log_{b}x$ is equivalent to $b^{y}=x$ (for $x\gt0$ and $b\gt0$, $b\ne1$). Therefore, $log_{10}10=1$, because $10^{1}=10$.
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.