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: 76

Answer

$x=\{ 1, 10 \}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \log x=\sqrt{\log x} ,$ square both sides of the equation. Then use concepts of solving quadratic equations to solve for the values of $x$. Finally, do checking of the solutions with the original equation. $\bf{\text{Solution Details:}}$ Squaring both sides of the equation above results to \begin{array}{l}\require{cancel} \left( \log x \right)^2=\left( \sqrt{\log x} \right)^2 \\\\ \left( \log x \right)^2=\log x .\end{array} Transposing $\log x$ and factoring the $GCF$ of the resulting expression result to \begin{array}{l}\require{cancel} \left( \log x \right)^2-\log x=0 \\\\ \log x(\log x-1)=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-1=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-1=0 \\\\ \log_{10} x=1 \\\\ x=10^1 \\\\ x=10 .\end{array} Upon checking, both $ x=\{ 1, 10 \} ,$ satisfy the original equation.
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