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 - Chapter Review Exercises: 50

Answer

$x = \sqrt {312.5}$

Work Step by Step

$\log_5 2x + \log_5 x = 4$ $\log_5 2x^{2} = 4$ $5^{4} = 2x^{2}$ $625 = 2x^{2}$ $x^{2} = 312.5$ $x = ±\sqrt {312.5}$. The negative value is not in the domain of $\log_5 x$. $x = \sqrt {312.5}$ Check: $\log_5 2x + \log_5 x \overset{?}{=} 4$ $\log_5 2(\sqrt {312.5}) + \log_5 (\sqrt {312.5}) \overset{?}{=} 4$ $\log_5 (2\sqrt {312.5}) + \log_5 (\sqrt {312.5}) \overset{?}{=} 4$ $\log_5 (2\sqrt {312.5})(\sqrt {312.5}) \overset{?}{=} 4$ $\log_5 625 \overset{?}{=} 4$ $5^{4}\overset{?}{=} 625$ $625 = 625$
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