Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 4 - Review - Test - Page 391: 4

Answer

(a) $10^{\log36}=36$ (b) $\ln{e^3}=3$ (c) $\log_3\sqrt{27}=\frac{2}{3}$ (d) $\log_280-\log_210=3$ (e) $\log_84=\frac{3}{2}$ (f) $\log_64+log_69=2$

Work Step by Step

A few remainder: $\log_mm^n=n$ $\log_ma-\log_mb=\frac{\log_ma}{\log_mb}$ $\log_ma+\log_mb=\log_m(a\times b)$ --- (a) $10^{\log36}=36^{\log10}=36^1=36$ Note, $\log{m}$ stands for logarithm with base $10$. Base $10$ is often omitted. (b) $\ln{e^3}=3$ $ln$ stands for logarithm with base $e$ (c) $\log_3\sqrt{27}=\log_33^{\frac{2}{3}}=\frac{2}{3}$ (d) $\log_280-\log_210=\log_2\frac{80}{10}=\log_28=\log_22^3=3$ (e) $\log_84=\log_88^{\frac{3}{2}}=\frac{3}{2}$ (f) $\log_64+log_69=\log_6(4\times9)=\log_636=\log_66^2=2$
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