Logarithm's basics

$\begin{aligned} \log_4 \sqrt[3]{8^{3+\frac{1}{3}\log_2{x}}} & = 4 \\ \sqrt[3]{8^{3+\frac{1}{3}\log_2{x}}} & = 4^4 \\ \left( 2^3 \right)^{\frac{1}{3}\left(3+\frac{1}{3}\log_2{x}\right)} & = 2^8 \\ 2^{3+\frac{1}{3}\log_2{x}} & = 2^8 \\ 3+\frac{1}{3}\log_2{x} & = 8 \\ \frac{1}{3}\log_2{x} & = 5 \\ \log_2{x} & = 15 \\ \Rightarrow x & = 2^{15} \\ \\ \Rightarrow a + b = \boxed{17} \end{aligned}$

$\begin{aligned}\begin{aligned}\log_4\sqrt[3]{8^{3+\frac13\log_2x}}=&\ 4\\\log_4{\left(8^{3+\frac13\log_2x}\right)^{\frac13}}=&\ 4\log_4{4}\\\log_4{8^{1+\frac19\log_2x}}=&\ \log_4{4^4}\\8^{1+\frac19\log_2x}=&\ 4^4\\(2^3)^{1+\frac19\log_2x}=&\ (2^2)^4\\2^{3+\frac13\log_2x}=&\ 2^8\\3+\frac13\log_2x=&\ 8\\\frac13\log_2x=&\ 5\\\log_2x=&\ 15\log_22\\\log_2x=&\ \log_2{2^{15}}\\x=&\ 2^{15}\end{aligned}\end{aligned}$

Thus, $a+b=2+15=\boxed{17}$

log [8^(3+(log x )/3log2)]^(1/3)/log4=4

[8^(3+(log x )/3log2)]^(1/3)=4^4

[8^(3+(log x )/3log2)]=4^12

3+(log x )/3log2=(log 4^12)/(log 8)

3+(log x )/3log2=8

(log x )/3log2=5

(log x)/(log 2)=15

x=2^15

a=2, b=15

Therefore, a+b=17

Know exponents? Then you got it. Easy one but well-constructed problem.