Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 3 - Section 3.4 - Exponential and Logarithmic Equations - Exercise Set - Page 489: 61

Answer

Solution set = $\displaystyle \{\frac{e^{4}}{2}\}.$ $ x\approx 27.30$

Work Step by Step

For the equation to be defined, logarithms must have positive arguments $2x \gt 0$ $ x \gt 0 \qquad (*)$ This is the condition eventual solutions must satisfy. $ 5\ln 2x=20\qquad $... divide with $5$ $\ln 2x=4\qquad $... apply $ e^{(...)}$, the inverse of $\ln $ $ e^{\ln 2x}=e^{4}$ $ 2x=e^{4}\qquad $... divide with $2$ $ x=\displaystyle \frac{e^{4}}{2}\qquad $... round to 2 decimal places $ x\approx 27.30$ (satisfies the condition (*), and is thus a valid solution.)
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