Intermediate Algebra: Connecting Concepts through Application

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

Chapter 6 - Logarithmic Functions - 6.6 Solving Logarithmic Equations - 6.6 Exercises: 41

Answer

$p \approx -0.349$

Work Step by Step

$\log (2p+1) = -0.5$ $10^{-0.5} = 2p + 1$ $10^{-0.5} - 1 = 2p$ $p = \frac{10^{-0.5}-1}{2}$ $p = -0.341886...$ $p \approx -0.349$ Check: $\log_{10} (2(\frac{10^{-0.5}-1}{2})+1) \overset{?}{=} -0.5$ $\log_{10} (10^{-0.5}-1+1) \overset{?}{=} -0.5$ $\log_{10} (10^{-0.5}) \overset{?}{=} -0.5$ $-0.5\log_{10} (10) \overset{?}{=} -0.5$ $-0.5(1) \overset{?}{=} -0.5$ $-0.5 = -0.5$
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