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: 23

Answer

$x\approx5.879$

Work Step by Step

Using the properties of logarithms, the given equation, $ \ln e^{0.45x}=\sqrt{7} $ is equivalent to \begin{align*}\require{cancel} 0.45x\ln e&=\sqrt{7} &(\text{use }\log_b x^y=y\log_b x) \\ 0.45x(1)&=\sqrt{7} &(\text{use }\ln e=1) \\ 0.45x&=\sqrt{7} .\end{align*} Using the properties of equality, the equation above is equivalent to \begin{align*}\require{cancel} \dfrac{\cancel{0.45}x}{\cancel{0.45}}&=\dfrac{\sqrt{7}}{0.45} \\\\ x&=\dfrac{\sqrt{7}}{0.45} \\\\ x&\approx5.879 .\end{align*} Hence, the solution to the equation $ \ln e^{0.45x}=\sqrt{7} $ is $ x\approx5.879 $.
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