Functions Modeling Change: A Preparation for Calculus, 5th Edition

Published by Wiley
ISBN 10: 1118583191
ISBN 13: 978-1-11858-319-7

Chapter 5 - Logarithmic Functions - Review Exercises and Problems for Chapter Five - Page 227: 48


$\log 25 \approx 1.4$

Work Step by Step

Use the fact that $(a^b)^n=a^{bn}$ to simplify the expression to $25=5^2=(10^{0.7})^2=10^{1.4}$. Therefore, $$\log 25=\log 10^{1.4}$$ Since $\log 10^a=a$ for all positive $a$, $\log 10^{1.4}\approx1.4$
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