Answer
$x=4$
Work Step by Step
$\log_{5}x+\log_{5}(x+1)=\log_{5}20$
Combine the logarithms on the left side:
$\log_{5}x(x+1)=\log_{5}20$
$\log_{5}(x^{2}+x)=\log_{5}20$
Since log is one to one, this equation becomes:
$x^{2}+x=20$
Subtract $20$ from each side:
$x^{2}+x-20=0$
Solve by factoring:
$(x-4)(x+5)=0$
We get two solutions, which are:
$x=4$ and $x=-5$
If we check our answers, we'll see that $x=-5$ is not a solution to this equation because $\log_5 (-5)$ is not defined.
.
So our final answer is $x=4$.
