An algebra problem by Aly Ahmed

$\begin{aligned} 60^m & = 5 & \small \blue{\text{Given}} \\ \frac {60}{60^m} & = \frac {60}5 \\ 60^{1-m} & = 12 & \small \blue{\text{Raise both sides to a power of }\frac 1{1-m}} \\ \implies 12^{\frac 1{1-m}} & = 60 & \small \blue{\text{Raise both sides to a power of }n} \\ 12^{\frac n{1-m}} & = 60^n = \boxed 3 & \small \blue{\text{It is given that }60^n = 3} \end{aligned}$

From the given equations we get $m=\dfrac{\ln 5}{\ln 60}\implies 1-m=\dfrac{\ln 12}{\ln 60},$

$n=\dfrac{\ln 3}{\ln 60}$

$\implies \dfrac{n}{1-m}=\dfrac{\ln 3}{\ln 12}$

$\implies 12^{\frac{n}{1-m}}=\boxed 3$ .