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 - Page 447: 77


$x=\{ 1,100 \}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \log x^2=(\log x)^2 ,$ use the properties of logarithms. Then use concepts of solving quadratic equations to solve the values of $x.$ Finally, do checking of the solutions with the original equation. $\bf{\text{Solution Details:}}$ 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} 2\log x=(\log x)^2 .\end{array} Using the properties of equality and factoring the $GCF,$ the equation above is equivalent to \begin{array}{l}\require{cancel} 0=(\log x)^2-2\log x \\\\ (\log x)^2-2\log x=0 \\\\ \log x(\log x-2)=0 .\end{array} Equating each factor to $0$ (Zero Product Property), the solutions are \begin{array}{l}\require{cancel} \log x=0 \\\\\text{OR}\\\\ \log x-2=0 .\end{array} Since $\log_by=x$ is equivalent to $y=b^x$, the equations above, in exponential form, are equivalent to \begin{array}{l}\require{cancel} \log x=0 \\\\ \log_{10} x=0 \\\\ x=10^0 \\\\ x=1 \\\\\text{OR}\\\\ \log x-2=0 \\\\ \log_{10} x=2 \\\\ x=10^2 \\\\ x=100 .\end{array} Upon checking, both $ x=\{ 1,100 \} ,$ satisfy the original equation.
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