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 489: 58


$x\approx -1.98$

Work Step by Step

RECALL: $\log_b{x}=y \longrightarrow b^y=x$ Write the equation in exponential form using the rule above to obtain: $7^{-2}=x+2 \\\frac{1}{7^2}=x+2 \\\frac{1}{49}=x+2$ Solve for $x$ to obtain: $\begin{array}{ccc} &\frac{1}{49}-2 &= &x+2-2 \\&\frac{1}{49} -1\frac{49}{49}&=&x \\&-1\frac{48}{49} &= &x \\&x &\approx &-1.98\end{array}$
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