Answer
Solution set = $\{3\}$
Work Step by Step
For the equation to be defined, logarithms must have positive arguments
$\begin{array}{ll}
x-1 \gt 0, & \\
x \gt 1 & (*)
\end{array}$
$(*)$ is the condition eventual solutions must satisfy.
$ 3\log_{2}(x-1)=5-\log_{2}4\qquad $... LHS: power rule, RHS: $\log_{a}a^{n}=n $
$\log_{2}(x-1)^{3}=\log_{2}2^{5}-\log_{2}2^{2}\qquad $ ... RHS: quotient rule
$\displaystyle \log_{2}(x-1)^{3}=\log_{2}\frac{2^{5}}{2^{2}}\qquad $ ... if $\log_{b}M=\log_{b}N,$ then M=N
... logarithmic functions are one to one,
$(x-1)^{3}=2^{3}$
... when n is odd, $ x^{n}$ is one to one,
$ x-1=2$
$ x=3\qquad $ ... satisfies (*)
Solution set = $\{3\}$
