## College Algebra 7th Edition

Published by Brooks Cole

# Chapter 4, Exponential and Logarithmic Functions - Section 4.5 - Exponential and Logarithmic Functions - 4.5 Exercises - Page 404: 65

#### Answer

$x=6$

#### Work Step by Step

$\log_{9}(x-5)+\log_{9}(x+3)=1$ $\log_{9}[(x-5)(x+3)]=1$ $(x-5)(x+3)=9^{1}$ $x^2+3x-5x-15=9$ $x^2-2x-15-9=0$ $x^{2}-2x-24=0$ $(x-6)(x+4)=0$ $(x-6)=0$ or $(x+4)=0$ $x=6$ or $x=-4$ However, $x=-4$ does not work in the original equation because we can't take the log of a negative number ($\log_{9}(-4-5)$=undefined). So we throw this solution out.

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