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: 102

Answer

The solution set is $\{6,-2\}.$

Work Step by Step

$ 3^{x^{2}-12}=9^{2x}\quad $ ... recognize $9$ as $3^{2}$ $ 3^{x^{2}-12}=(3^{2})^{2x}\quad $ ... apply $(a^{m})^{n}=a^{mn}$ $ 3^{x^{2}-12}=3^{4x}\qquad $ ... exponential functions are one to one. $ x^{2}-12=4x $ $ x^{2}-4x-12=0$ Factor by finding factors of -12 whose sum is -4 $(x-6)(x+2)=0$ Apply the zero product property $ x-6=0$ or $ x+2=0$ The solution set is $\{6,-2\}.$
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