College Algebra (10th Edition)

Published by Pearson

Chapter 6 - Section 6.3 - Exponential Functions - 6.3 Assess Your Understanding - Page 436: 83

Answer

$\dfrac{1}{49}$

Work Step by Step

RECALL: (1) $a^{-m} = \dfrac{1}{a^m}$ (2) $a^{mn}= (a^m)^n$ Use rule (1) above to obtain: $4^{-2x} = \dfrac{1}{4^{2x}}$ Use rule (2) above to obtain: $\dfrac{1}{4^{2x}} = \dfrac{1}{(4^x)^2}$ Thus, $4^{-2x} = \dfrac{1}{(4^x)^2}$ With $4^x=7$, the expression above simplifies to: $=\dfrac{1}{7^2} \\=\dfrac{1}{49}$

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