Merry New Year! 2

We have, $\sqrt{\dfrac{\log _{ 2015 }{ 2016 }}{\log _{ 2016 }{ 2015 }}} \\ =\sqrt{\dfrac{\frac{\log 2016}{\log 2015}}{\frac{\log 2015}{\log 2016}}} =\sqrt{\dfrac{(\log 2016)^2}{(\log 2015)^2}} \\ =\dfrac{\log2016}{\log 2015} \\ \therefore \color{#D61F06}{A}=\log_{2015}{2016} \\ \implies 2015^{\color{#D61F06}{A}}=2015^{\log_{2015}{2016}} \\ =\boxed{2016}.$

Note : $\large{\log_{\color{#20A900}{a}}{\color{#3D99F6}{b}}=\dfrac{\log \color{#3D99F6}{b}}{\log \color{#20A900}{a}}} \quad\quad\quad[\text{Base-changing rule}] \\ \large{\color{#20A900}{a}^{\log_{\color{#20A900}{a}}{\color{#3D99F6}{b}}}=\color{#3D99F6}{b}}$

$\large \color{#D61F06}{A} ~=~ \sqrt{\frac{\log_{2015}{2016}}{\log_{2016}{2015}}}$

$\large \log_{2016}{2015} ~=~ \frac{\log_{2015}{2015}}{\log_{2015}{2016}} ~=~ \frac{1}{\log_{2015}{2016}}$

$\large \color{#D61F06}{A} ~=~ \sqrt{\frac{\log_{2015}{2016}}{\frac{1}{\log_{2015}{2016}}}}$

$\large \color{#D61F06}{A} ~=~ \sqrt{\log_{2015}{2016} \cdot \log_{2015}{2016}} ~=~ \sqrt{\left(\log_{2015}{2016}\right)^2}$

$\large \color{#D61F06}{A} ~=~ \log_{2015}{2016}$

$\large 2015^{\color{#D61F06}{A}} ~=~ 2015^{\log_{2015}{2016}} ~=~ 2016$

The key to the solution lies in the fact that log with base b and argument a is equal to 1/(log base a argument b). This is a direct result of the base change formula. Sorry I don't know how to use Latex.

The answer is simple when we change the base lets say to a common base b

We get log {a} 2016/log {a}2016

This says A=log_{2015}2016

So we know x^log_{x}y = y it's prove is easy too