College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 4 - Section 4.5 - Exponential and Logarithmic Equations - 4.5 Exercises - Page 447: 53



Work Step by Step

Using the Product Property of logarithms, the given expression, $ \log(x+25)=\log(x+10)+\log4 $ is equivalent to \begin{array}{l}\require{cancel} \log(x+25)=\log[4(x+10)] .\end{array} Since the bases of the logarithms are the same, then these can be dropped. Hence, \begin{array}{l}\require{cancel} x+25=4(x+10) .\end{array} Using the properties of equality, the solution to the equation above is \begin{array}{l}\require{cancel} x+25=4x+40 \\\\ x-4x=40-25 \\\\ -3x=15 \\\\ x=\dfrac{15}{-3} \\\\ x=-5 .\end{array} Upon checking, $ x=-5 $ satisfies the original equation.
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