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: 94

Answer

Solution set = $\{ 1 \}$

Work Step by Step

$3^{x+2}\cdot 3^{x}=81\quad \left\{\begin{array}{ll} \mathrm{L}\mathrm{H}\mathrm{S}: & a^{n}\cdot a^{m}=a^{n+m}\\ \mathrm{R}\mathrm{H}\mathrm{S}: & 125=5^{3} \end{array}\right.$ $3^{(x+2)+x}=3^{4}$ $ 3^{2x+2}=3^{4}\qquad $ ... exponential functions are one to one. $2x+2=4$ $2x=2$ $ x=1$ Solution set = $\{ 1 \}$
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