Answer
$3$
Work Step by Step
The domain of the variable requires that $x-1>0$.
This means that $x>1$.
Recall:
$\log_a M^r = r \log_a M$
Using the rule above gives:
$3 \log_2 (x-1) = \log_2 (x-1)^3$
Thus, the given equation is equivalent to:
$\log_2{(x-1)^3}+\log_2{4}=5$
Recall further that:
$\log_a (MN) = \log_a M+\log_a N$
Hence, the equation above is equivalent to:
$\log_2 4(x-1)^3=5$
Since $y = \log_a b \text{ is equivalent to } b=a^y$, then
$\log_2 4(x-1)^3 = 5 \text{is equivalent to } 4(x-1)^3 = 2^5$
Solve the equation to obtain:
$4(x-1)^3 = 32$
$(x-1)^3 = \dfrac{32}{4}$
$(x-1)^3 = 8$
$x-1 = \sqrt[3]{8}$
$x-1 = 2$
$x = 2+1$
$x=\boxed{3}$
