Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 12 - Exponential Functions and Logarithmic Functions - Connecting: The Concepts - Exercises - Page 824: 4

Answer

$x=5$ or $x=-5$

Work Step by Step

By the principle of logarithmic equality, $\log_{3}(x^{2}+1)=\log_{3}26\quad \Rightarrow\quad x^{2}+1=26$ ... subtract 1 from both sides... $x^{2}=25$ $x=\pm 5$ (The original equation is defined for either $x=5$ or $x=-5$, so they are valid solutions.)
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