Reach for the Summit - M-S1-A2

From $\log_2 a = \dfrac {a+1}b$ $\implies a = 2^\frac {a+1}b$ . Then we have:

$\begin{aligned} 4^a - 3a^b & = 16 \\ 2^{2a} - 3 \left(2^\frac {a+1}b \right)^b & = 16 \\ 2^{2a} - 6 \left(2^a \right) & = 16 \\ \left(2^a\right)^2 - 6 \left(2^a \right) - 16 & = 0 \\ \left(2^a - 8 \right) \left(2^a + 2\right) & = 0 \\ 2^a & = 8 & \small \blue{\text{Note that} 2^a = -2 \text{ has no real solution.}} \\ \implies a & = 3 \\ \implies 4^3 - 3a^b & = 16 \\ 3 a^b & = 4^3 - 16 = 48 \\ \implies a^b = \boxed{16} \end{aligned}$

From the second equation, $a^b=2^{a+1}=2\times 2^a=2x$ (say)

Then the first equation becomes

$x^2-6x-16=0$ , whose solutions are $8,-2$

Since $x=2^a$ can't be negative, therefore $2^a=8\implies a=3$ , and $a^b=2^4=\boxed {16}$ .