Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 4 - Review - Exercises - Page 389: 59



Work Step by Step

$2^{3x-5}=7$ Apply $\log$ to both sides of the equation: $\log2^{3x-5}=\log7$ Take $3x-5$ to multiply in front of its $\log$: $(3x-5)\log2=\log7$ Solve for $x$: $3x-5=\dfrac{\log7}{\log2}$ $3x=\dfrac{\log7}{\log2}+5$ $x=\dfrac{\log7}{3\log2}+\dfrac{5}{3}\approx2.60$
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