College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 6 - Section 6.6 - Logarithmic and Exponential Equations - 6.6 Assess Your Understanding - Page 466: 59



Work Step by Step

...$3^{x+1}=3^{x}\cdot 3$ ... Substitute $t=3^{x}$ ... the equation becomes $t^{2}+3t-4=0 \quad $... factors of $-4$ whose sum is $+3$... $(t-1)(t+4)=0$ t is either $-4$ or $1$. $t=3^{x}=-4$ cannot be a solution as $3^{x}$ is never negative. $t=3^{x}=1$ $3^{x}=3^{0} \quad $...Apply$: \quad $If $a^{u}=a^{v},$ then $u=v$ $x=0$ The solution set is $\{0\}$
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