Intermediate Algebra (12th Edition)

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

Chapter 9 - Review Exercises - Page 638: 50

Answer

$\left\{18.310\right\}$

Work Step by Step

Taking the natural logarithm of both sides, the given equation, $ e^{0.06x}=3 $ is equivalent to \begin{align*}\require{cancel} \ln e^{0.06x}&=\ln 3 .\end{align*} Using the properties of logarithms, the equation above is equivalent to \begin{align*}\require{cancel} 0.06x(\ln e)&=\ln 3 &(\text{use }\log_b x^y=y\log_b x) \\ 0.06x(1)&=\ln 3 &(\text{use }\ln e=\log_e e=1) \\ 0.06x&=\ln 3 .\end{align*} Using the properties of equality, the equation above is equivalent to \begin{align*}\require{cancel} \dfrac{\cancel{0.06}x}{\cancel{0.06}}&=\dfrac{\ln 3}{0.06} \\\\ x&=\dfrac{\ln 3}{0.06} .\end{align*} Using a calculator, the approximate value of the logarithmic expression above is \begin{align*} \log3&\approx1.09861 .\end{align*} Substituting the approximate values in $ x=\dfrac{\ln 3}{0.06} $, then \begin{align*} x&\approx\dfrac{1.09861}{0.06} \\\\ x&\approx18.310 .\end{align*} Hence, the solution set to the equation $ e^{0.06x}=3 $ is $ \left\{18.310\right\} $.
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