(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.