Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 9 - Review Exercises - Page 750: 31



Work Step by Step

Note that $\dfrac{1}{25}=\dfrac{1}{5^2} = 5^{-2}$. Write $\dfrac{1}{25}$ as $5^{-3}$ to obtain: $=\log_5{(5^{-3})}$ Use the rule "$\log_b{(b^x)}=x$" to obtain: $\log_5{(5^{-2})} = -2$
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