Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 9 - Section 9.5 - Exponential and Logarithmic Equations - Exercise Set: 52

Answer

$x=\dfrac{-97}{49}$

Work Step by Step

RECALL: $\log_b{x}= y \longrightarrow b^y=x$ Use the rule above to obtain: $7^{-2}=x+2 \\\dfrac{1}{7^2}=x+2 \\\dfrac{1}{49}=x+2$ Subtract $2$ on both sides of the equation to obtain: $\dfrac{1}{49}-2=x \\\dfrac{1}{49} - \dfrac{98}{49}=x \\\dfrac{-97}{49}=x$
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