## College Algebra 7th Edition

$x=4$
$\log_{5}x+\log_{5}(x+1)=\log_{5}20$ $\log_{5}(x(x+1))=\log_{5}20$ $\log_{5}(x^{2}+x)=\log_{5}20$ $x^{2}+x=20$ $x^{2}+x-20=0$ $(x+5)(x-4)=0$ $(x+5)=0$ 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.