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 - Page 543: 48

Answer

$x = 9998$

Work Step by Step

$\log (10x+20) = 5$ $\log_{10} (10x+20) = 5$ $10^{5} = 10x+20$ $100000 = 10x+20$ $99980 = 10x$ $x = 9998$ Check: $\log (10(9998)+20) \overset{?}{=} 5$ $\log (99980+20) \overset{?}{=} 5$ $\log (100000) \overset{?}{=} 5$ $10^{5} \overset{?}{=} 100000$ $100000 = 100000$
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