Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 5 - Exponential Functions - 5.2 Solving Equations Using Exponent Rules - 5.2 Exercises: 101

Answer

1. Exponential equation 2. $x = 7$

Work Step by Step

1. Exponential equation For example: $7^{x} + 3 = 0$ is an exponential equation. (The variable is the exponent.) $x^{7} + 3 =0 $ is a power equation. (The variable is raised to the power of a number.) 2. Solve $5(\frac{1}{2})^{x} - \frac{2}{32} = 3(\frac{1}{2})^{x} - \frac {3}{64}$ $5(\frac{1}{2})^{x} - 3(\frac{1}{2})^{x} = \frac{2}{32} - \frac{3}{64}$ $(5-3)((\frac{1}{2})^{x}) = \frac{1}{64}$ $2((\frac{1}{2})^{x}) = \frac{1}{64}$ $(\frac{1}{2})^{x} = \frac{1}{128}$ $2^{-x} = 128^{-x}$ $2^{-x} = 2^{-7}$ $-x = -7$ $x = 7$ Check: $5(\frac{1}{2})^{7} - \frac{2}{32} \overset{?}{=} 3(\frac{1}{2})^{7} - \frac {3}{64}$ $5(\frac{1}{128}) - \frac{2}{32} \overset{?}{=} 3(\frac{1}{128}) - \frac {3}{64}$ $(\frac{5}{128}) - \frac{2}{32} \overset{?}{=} (\frac{3}{128}) - \frac {3}{64}$ $-\frac{3}{128} = -\frac{3}{128}$
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