A log to roll over

$\large \log_{2\sqrt{3}}(1728) = x \Longrightarrow 2^{x}3^{\frac{x}{2}} = 12^{3} = 2^{6}3^{3} \Longrightarrow x = 6$

by the Fundamental Theorem of Arithmetic, (FTA).

$\large \log_{2\sqrt[5]{63}}(2016) = y \Longrightarrow (2\sqrt[5]{3^{2}*7})^{y} = 2016 \Longrightarrow 2^{y}3^{\frac{2y}{5}}7^{\frac{y}{5}} = 2^{5}3^{2}7 \Longrightarrow y = 5$ ,

again by the FTA. Thus $x + y = 6 + 5 = \boxed{11}$ .

Just write it as:

$\Large \log_{\sqrt{4\cdot 3}}(12^3) + \log_{\sqrt[5]{32\cdot 63}}(2016)$

$\color{#D61F06}{(\log_{a^{\frac 1p}} b=p\log_a b)}~~a,b>0,a\neq 1,p\neq 0$

$\Large 2\log_{12}(12^3) + 5\log_{2016}(2016)$

$\color{#D61F06}{(\log_a a^c =c)}~~a,c>0, a\neq 1$

$=6+5=\boxed{11}$

$\dots = \frac{\log 1728}{\log 2\sqrt 3} + \frac{\log 2016}{\log 2\sqrt[5]{63}} \\ = \frac{6\log 2 + 3 \log 3}{\log 2 + \tfrac12\log 3} + \frac{5\log 2 + 2\log 3 + \log 7}{\log 2 + \tfrac25\log 3 + \tfrac15\log 7} \\ = \frac{3(2 \log 2 + \log 3)}{\tfrac12(2\log 2 + \log 3)} + \frac{5\log 2 + 2\log 3 + \log 7}{\tfrac15(5\log 2 + 2\log 3 + \log 7)} \\ = \frac{3}{1/2} + \frac{1}{1/5} = 6 + 5 = \boxed{11}.$

the question is same as

(log (1728)/log (12^(1/3))) + (log (2016) / log ( 2016^(1/5))

1728 = 12^3

let log 1728= a and log 2016 = b

hence ( 3*a)/(a/2) + (b)/(b/5) = 6 + 5 = 11