College Algebra (11th Edition)

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

Chapter 4 - Review Exercises: 58

Answer

$x=\pm6$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \log_3(x^2-9)=3 ,$ convert to exponential form. Then use the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ Since $\log_by=x$ is equivalent to $y=b^x$, the equation above, in exponential form, is equivalent to \begin{array}{l}\require{cancel} x^2-9=3^3 \\\\ x^2-9=27 .\end{array} Using the properties of equality to isolate the variable results to \begin{array}{l}\require{cancel} x^2=27+9 \\\\ x^2=36 \\\\ \sqrt{x^2}=\pm\sqrt{36} \\\\ x=\pm\sqrt{(6)^2} \\\\ x=\pm6 .\end{array}
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