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: 17

Answer

$x=-\dfrac{1}{4}$

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) above to obtain $4^{x}=\dfrac{1}{2^{\frac{1}{2}}}.$ Use rule (iii) above to obtain $4^x=2^{-\frac{1}{2}}.$ Write 4 as a power of 2 to obtain $(2^2)^x = 2^{-\frac{1}{2}} \\2^{2x} =2^{-\frac{1}{2}}.$ Use rule (i) above to obtain $2x=-\frac{1}{2}.$ Solve the equation: $\begin{array}{ccc} &2x &= &-\frac{1}{2} \\&\frac{1}{2} (2x) &= &-\dfrac{1}{2} \cdot \dfrac{1}{2} \\&x &= &-\dfrac{1}{4} \\\end{array}.$
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