Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 9 - Section 9.6 - Exponential and Logarithmic Equations; Further Applications - 9.6 Exercises - Page 630: 18

Answer

$x\approx566.866$

Work Step by Step

Taking the natural logarithm of both sides, the given equation, $ e^{0.006x}=30 $ is equivalent to \begin{align*}\require{cancel} \ln e^{0.006x}=\ln30 .\end{align*} Using the properties of logarithms, the equation above is equivalent to \begin{align*}\require{cancel} 0.006x(\ln e)&=\ln30 &(\text{use }\log_b x^y=y\log_b x) \\ 0.006x(1)&=\ln30 &(\text{use }\ln e=1) \\ 0.006x&=\ln30 .\end{align*} Using the properties of equality, the equation above is equivalent to \begin{align*}\require{cancel} \dfrac{\cancel{0.006}x}{\cancel{0.006}}&=\dfrac{\ln30}{0.006} \\\\ x&=\dfrac{\ln30}{0.006} \\\\ x&\approx566.866 .\end{align*} Hence, the solution to the equation $ e^{0.006x}=30 $ is $ x\approx566.866 $.
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