College Algebra 7th Edition

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

Chapter 4, Exponential and Logarithmic Functions - Chapter 4 Review - Concept Check - Page 425: 69



Work Step by Step

We solve: $\log_{3}(x-8)+\log_{3}x=2$ $\log_{3}((x-8)*x)=2$ $\log_{3}(x^2-8x)=2$ $3^2=x^2-8x$ $ x^{2}-8x-9=0$ $(x-9)(x+1)=0$ $(x-9)=0$ or $(x+1)=0$ $x=9$ or $x=-1$ However, $x=-1$ is not allowed in the original equation because we can't take the log of a negative number (undefined). So we throw this solution out.
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