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 488: 48

Answer

Solution set = $\displaystyle \{\frac{\ln 3}{\ln 2}\}.$ $ x\approx 1.58$

Work Step by Step

Recognize that $2^{2x}=(2^{x})^{2}$. Introduce a new variable t: $ t=2^{x}.$ $ t \gt 0 $, because $2^{x}$ is always positive. We solve $ t^{2}+t-12=0$ Factor by finding factors of $-12$ whose sum is $+1$: $(t+4)(t-3)=0$ $ t+4=0$ or $ t-3=0$ $ t=-4$ or $ t=3$ $ t=-4$ is discarded, as $ t \gt 0 $, leaving $ t=3\qquad $... bring back x $ 2^{x}=3\qquad $... apply ln( ) $\ln 2^{x}=\ln 3\qquad $... apply $\log_{b}a^{p}=p\log_{b}a $ $ x\ln 2=\ln 3\qquad $... divide with $\ln 2$ $ x=\displaystyle \frac{\ln 3}{\ln 2}\qquad $... round to 2 decimal places $ x\approx 1.58$
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