College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 4 - Section 4.4 - Evaluating Logarithms and the Change-of-Base Theorem - 4.4 Exercises: 14

Answer

$-2$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the properties of logarithms to simplify the given expression, $ \log 0.01 .$ $\bf{\text{Solution Details:}}$ The given expression is equivalent to \begin{array}{l}\require{cancel} \log \dfrac{1}{100} \\\\= \log \dfrac{1}{10^2} .\end{array} Using the Quotient Rule of Logarithms, which is given by $\log_b \dfrac{x}{y}=\log_bx-\log_by,$ the expression above is equivalent \begin{array}{l}\require{cancel} \log 1-\log 10^2 .\end{array} Using the Power Rule of Logarithms, which is given by $\log_b x^y=y\log_bx,$ the expression above is equivalent \begin{array}{l}\require{cancel} \log 1-2\log 10 .\end{array} Since $\log_b1=0$ and $\log_b b=1,$ the expression above is equivalent \begin{array}{l}\require{cancel} \log 1-2\log_{10} 10 \\\\= 0-2(1) \\\\= 0-2 \\\\= -2 .\end{array}
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