(a) 3 log${}_2$ x = y $\to$ 2${}^y$ = x${}^3$.............(1)
Similarly, log${}_2$ 4x = y + 4 can be written as 2${}^{y+4}$ = 4x............(2)
Substituting for 2y in equation (2) to have 16x${}^3$ = 4x and to form a cubic equation 16x${}^3$ - 4x = 0.
Then, using factorizing to have 4x(4x${}^2$ - 1) = 0 and solving for x to get x = 0 or x = $\frac{1}{2}$ or $\frac{1}{2}$
Next, substituting for x in equation (1), when x = 0, y has no solution.
Also, when x = -$\frac{1}{2}$ y has no solution and when x = $\frac{1}{2}$, 2${}^y$ = ($\frac{1}{2}$)${}^3$ = 2${}^{-3}$
Therefore, y = -3
(b). Substitute to have -3*5 = (-3)${}^2$ - 2(-3) (5) + 5${}^2$ = 2${}^n$ and when simplified to get 64 = 2${}^n$. However, 2${}^6$ = 2${}^n$ so that n = 6.