College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 4, Exponential and Logarithmic Functions - Chapter 4 Review - Concept Check - Page 425: 60


$x \approx 1.58$

Work Step by Step

Take the common log of both sides to obtain: $\log{(10^{6-3x})} = \log{18}$ Use the rule $\log{(a^n)} = n\cdot\log{a}$ to obtain: $(6-3x)\log{10} = \log{18}$ Divide $\log{10}$ to both sides of the equation to obtain: $\dfrac{(6-3x)\log{10}}{\log{10}} = \dfrac{\log{18}}{\log{10}} \\6-3x=\dfrac{\log{18}}{\log{10}}$ Subtract $6$ on both sides of the equation to obtain: $-3x=\dfrac{\log{18}}{\log{10}}-6$ Divide $-3$ on both of the equation to obtain: $x = \dfrac{\frac{\log{18}}{\log{10}}-6}{-3}$ Use a calculator to obtain: $x \approx 1.58$
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