College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 4, Exponential and Logarithmic Functions - Section 4.5 - Exponential and Logarithmic Functions - 4.5 Exercises: 50

Answer

$x=4$

Work Step by Step

$\log_{5}x+\log_{5}(x+1)=\log_{5}20$ $\log_{5}(x(x+1))=\log_{5}20$ $\log_{5}(x^{2}+x)=\log_{5}20$ $x^{2}+x=20$ $x^{2}+x-20=0$ $(x+5)(x-4)=0$ $(x+5)=0$ or $(x-4)=0$ $x=-5$ or $x=4$ But $x=-5$ does not work in the original equation because we can't take a log of a negative number (undefined). So we throw this solution out.
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