Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 3 - Review Exercises - Page 513: 74


Solution set = $\{23\}$.

Work Step by Step

$\log_{4}(3x-5)=3$ Logarithms are defined only for positive arguments, so any solution must satisfy $3x-5 \gt 0\Rightarrow\qquad x \gt 5/3\qquad(*)$ Write the RHS as $\log_{4}($....$)$ using the basic property $\log_{b}b^{x}=x $ $\log_{4}(3x-5)=\log_{4}4^{3}$ logarithmic functions are one-to-one; if $\log_{b}M=\log_{b}N $, then M=N $3x-5=4^{3}$ $ 3x-5=64\qquad $... add 5 $ 3x=69\qquad $... divide with 3 $ x=23$ which satisfies the condition (*), and is a valid solution. Solution set = $\{23\}$.
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