Answer
please see "step by step"
Work Step by Step
We use the Basic Logarithmic Properties
$\log_{b}b^{x}=x\qquad$ and $\qquad b^{\log_{b}x}=x$
in reducing an equation from logarithmic form to one that isn't.
The SECOND equation is of the form $\log_{3}(...)=\log_{3}(...) $
and we solve it by applying $3^{(..) }$to both sides. It then becomes
$ x-1=4\qquad$(...no longer logarithmic)
and its solution is
$x=5$
The FIRST equation does not have $\log_{3}(...)$ on the RHS.
So we use $\log_{b}b^{x}=x$, that is
$4=\log_{3}(3^{4})$, by which the equation becomes of the form $\log_{3}(...)=\log_{3}(...)$
$\log_{3}(x-1)=\log_{3}(3^{4})$
$... $we solve it by applying $3^{(..) }$to both sides$...$
$ x-1=3^{4}\qquad$(...no longer logarithmic)
$x-1=81$
$x=82$
