Answer
$x=4$
Work Step by Step
Using the properties of logarithms, the given equation, $\log(x-3)+\log(x+1)=\log5,$ is equivalent to
\begin{align*}\require{cancel}
\log[(x-3)(x+1)]&=\log5
&(\text{use }\log_b xy=\log_b x+\log_b y)
.\end{align*}
Since the logarithm on both sides of the equation have the same base, the arguments can be equated. That is,
\begin{align*}
(x-3)(x+1)&=5
.\end{align*}
In the form $ax^2+bx+c=0,$ the equation above is equivalent to
\begin{align*}
x^2-3x+x-3&=5
&(\text{use }(a+b)(c+d)=ac+bc+ad+bd)
\\
x^2-3x+x-3-5&=0
\\
x^2-2x-8&=0
&(\text{combine like terms})
\\
(x-4)(x+2)&=0
&(\text{use factoring of quadratic trinomials})
\end{align*}
Equating each factor to zero (Zero Product Property), the solutions to the equation above are
\begin{array}{l|r}
x-4=0 & x+2=0
\\
x=4 & x=-2
\end{array}
Check the solutions:
\begin{align*}
\text{If }x=4:&
\\&
\log(4-3)+\log(4+1)\overset{?}=\log5
\\&
\log1+\log5\overset{?}=\log5
\\&
0+\log5\overset{?}=\log5
&(\log_b 1=0)
\\&
\log5\overset{\checkmark}=\log5
\\\\
\text{If }x=-2:&
\\&
\log(-2-3)+\log(-2+1)\overset{?}=\log5
\\&
\log(-5)+\log(-1)\overset{?}=\log5
\\&
\text{undefined, since $x$ must be greater than $0$ in $\log x$.}
\end{align*}
Hence, the only solution to the equation $\log(x-3)+\log(x+1)=\log5
$ is $x=4$.
