Answer
$x=\dfrac{4}{3}$
Work Step by Step
Using the properties of logarithms, the given expression, $
\ln3+\ln (x-1)=0
,$ is equivalent to
\begin{array}{l}\require{cancel}
\ln[3(x-1)]=0
\\\\
\ln(3x-3)=0
\\\\
\log_e(3x-3)=0
.\end{array}
Since $y=b^x$ is equivalent to $\log_b y=x$, then the solution to the equation, $
\log_e(3x-3)=0
,$ is
\begin{array}{l}\require{cancel}
3x-3=e^0
\\\\
3x-3=1
\\\\
3x=1+3
\\\\
3x=4
\\\\
x=\dfrac{4}{3}
.\end{array}
Upon checking, $
x=\dfrac{4}{3}
$ satisfies the original equation.
