College Algebra (11th Edition)

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

Chapter 4 - Section 4.5 - Exponential and Logarithmic Equations - 4.5 Exercises: 89

Answer

$A=\dfrac{B}{x^C}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \log A=\log B-C\log x ,$ in terms of $ A ,$ use the properties of logarithms to simplify and isolate the needed variable. $\bf{\text{Solution Details:}}$ Using the Power Rule of Logarithms, which is given by $\log_b x^y=y\log_bx,$ the equation above is equivalent \begin{array}{l}\require{cancel} \log A=\log B-\log x^C .\end{array} Using the Quotient Rule of Logarithms, which is given by $\log_b \dfrac{x}{y}=\log_bx-\log_by,$ the equation above is equivalent \begin{array}{l}\require{cancel} \log A=\log \dfrac{B}{x^C} .\end{array} Since both sides are expressed in a logarithm with the same base, then the logarithm may be dropped. Hence, the equation above becomes \begin{array}{l}\require{cancel} A=\dfrac{B}{x^C} .\end{array}
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