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: 18

Answer

$x=-\dfrac{1}{6}$

Work Step by Step

RECALL: (i) If $a^M=a^N$, then $M=N.$ (ii) $\sqrt[n]{a} = a^{\frac{1}{n}}.$ (iii) $\dfrac{1}{a^m} = a^{-m}.$ Use rule (ii) to obtain $9^x=\dfrac{1}{3^{\frac{1}{3}}}.$ Use rule (iii) to obtain $9^x=3^{-\frac{1}{3}}.$ Write 9 as a power of 3 to obtain $(3^2)^x=3^{-\frac{1}{3}} \\3^{2x} = 3^{-\frac{1}{3}}.$ Use rule (i) above to obtain $2x=-\dfrac{1}{3}.$ Solve the equation to obtain $\dfrac{1}{2} \cdot 2x= -\dfrac{1}{3} \cdot \dfrac{1}{2} \\x = -\dfrac{1}{6}.$
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