College Algebra 7th Edition

Published by Brooks Cole

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

Answer

$x=9$

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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