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: 60



Work Step by Step

$10^{6-3x}=18$ Apply $\log$ to both sides of the equation: $\log10^{6-3x}=\log18$ Take $6-3x$ to multiply in front of its $\log$: $(6-3x)\log10=\log18$ Since $\log10=1$, this equation becomes: $6-3x=\log18$ $-3x=\log18-6$ $x=\dfrac{6}{3}-\dfrac{\log18}{3}$ $x=2-\dfrac{\log18}{3}\approx1.58$
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