[Algebra]
Problem: What are all values of $x$ such that

$$
\log _{5}(x+3)+\log _{5}(x-1)=1 ?
$$

Solution: Combining the logarithms,

$$
\begin{aligned}
\log _{5}(x+3)+\log _{5}(x-1) & =1 \\
\log _{5}((x+3)(x-1)) & =1 \\
\log _{5}\left(x^{2}+2 x-3\right) & =1 \\
x^{2}+2 x-3 & =5 \\
x^{2}+2 x-8 & =0 \\
(x+4)(x-2) & =0
\end{aligned}
$$

Therefore, $x=-4$ or $x=2$. Substituting the two values for $x$ back into the original equation, we see that $x=2$ works, but that $x=-4$ does not, since we cannot take the logarithm of a negative number.

Final answer: 2