Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 12 - Exponential Functions and Logarithmic Functions - 12.6 Solving Exponential Equations and Logarithmic Equations - 12.6 Exercise Set - Page 826: 97



Work Step by Step

$\log x=\ln x$ Write the RHS using change-of-base (to base 10) $\displaystyle \log x=\frac{\log x}{\log e}$ $\displaystyle \log x-\frac{\log x}{\log e}=0$ $\displaystyle \log x(1-\frac{1}{\log e})=0$ Since $(1-\displaystyle \frac{1}{\log e})\neq 0$, it must be that $\log x=0$. That is, $\quad x=1.$ ALTERNATIVELY, Graphing approach: Graph $y=\log x$ and $y=\ln x$ on the same coordinate system. The only intersection point is (1,0), thus, $x=1.$
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