## College Algebra 7th Edition

$x=5$
$\log x+\log(x-3)=1$ $\log[x(x-3)]=1$ $\log_{10}[x^2-3x]=1$ $x^{2}-3x=10^1$ $x^{2}-3x-10=0$ $(x+2)(x-5)=0$ $(x+2)=0$ or $(x-5)=0$ $x=-2$ or $x=5$ However, $x=-2$ does not work in the original equation because we can't take the log of a negative number (undefined). So we throw this solution out.