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: 13



Work Step by Step

RECALL: (i) If $a^M=a^N$, then $M=N.$ (ii) $(a^m)^n=a^{mn}.$ Note that $\dfrac{1}{27}=\dfrac{1}{3^3} = 3^{-3}$. Thus, the given equation can be written as $3^{1-x} = 3^{-3}.$ Use rule (i) above to obtain $1-x=-3 \\-x = -3-1 \\-x=-4 \\(-1)(-x)=-4(-1) \\x=4.$
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