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