Past, present, and everything in between

First, in order to transform the main expression into a more manipulable one, we can rationalize it:

$\sqrt[{{2^i}}]{{2015}} + \sqrt[{{2^i}}]{{1995}} \cdot \frac{{\sqrt[{{2^i}}]{{2015}} - \sqrt[{{2^i}}]{{1995}}}}{{\sqrt[{{2^i}}]{{2015}} - \sqrt[{{2^i}}]{{1995}}}} = \frac{{{{\left( {\sqrt[{{2^i}}]{{2015}}} \right)}^2} - {{\left( {\sqrt[{{2^i}}]{{1995}}} \right)}^2}}}{{\sqrt[{{2^i}}]{{2015}} - \sqrt[{{2^i}}]{{1995}}}} = \frac{{\sqrt[{{2^{i - 1}}}]{{2015}} - \sqrt[{{2^{i - 1}}}]{{1995}}}}{{\sqrt[{{2^i}}]{{2015}} - \sqrt[{{2^i}}]{{1995}}}}$

Easily we can observe that the new expression will let us to simplify factors while developing the Product. Thus:

$\prod\limits_{i = 1}^{{2^{2015}}} {\left( {\sqrt[{{2^i}}]{{2015}} + \sqrt[{{2^i}}]{{1995}}} \right)} = \frac{{2015 - 1995}}{{\sqrt {2015} - \sqrt {1995} }} \cdot \frac{{\sqrt {2015} - \sqrt {1995} }}{{\sqrt[4]{{2015}} - \sqrt[4]{{1995}}}} \cdot \frac{{\sqrt[4]{{2015}} - \sqrt[4]{{1995}}}}{{\sqrt[8]{{2015}} - \sqrt[8]{{1995}}}} \cdot \cdots \cdot \frac{{\sqrt[{{2^{{2^{2015}} - 1}}}]{{2015}} - \sqrt[{{2^{{2^{2015}} - 1}}}]{{1995}}}}{{\sqrt[{{2^{{2^{2015}}}}}]{{2015}} - \sqrt[{{2^{{2^{2015}}}}}]{{1995}}}}$

The numerator of any factor simplifies with the denominator of the preceeding one. Then, the result is:

$\prod\limits_{i = 1}^{{2^{2015}}} {\left( {\sqrt[{{2^i}}]{{2015}} + \sqrt[{{2^i}}]{{1995}}} \right)} = \frac{{2015 - 1995}}{{\sqrt[{{2^{{2^{2015}}}}}]{{2015}} - \sqrt[{{2^{{2^{2015}}}}}]{{1995}}}} = \frac{{20}}{{{{\left( {2015} \right)}^{\frac{1}{{{2^{{2^{2015}}}}}}}} - {{\left( {1995} \right)}^{\frac{1}{{{2^{{2^{2015}}}}}}}}}}$

So, the reciprocal would be $\frac{{{{\left( {2015} \right)}^{\frac{1}{{{2^{{2^{2015}}}}}}}} - {{\left( {1995} \right)}^{\frac{1}{{{2^{{2^{2015}}}}}}}}}}{{20}}$ and the requested values:

$\begin{array}{l} a = 2015\\ b = {2^{2015}}\\ c = 1995\\ d = 20 \end{array}$

Finally, substituing:

$\frac{{c{{\log }_2}\left( b \right) + 5a}}{{{{\log }_2}\left( {{b^d}} \right)}} = \frac{{1995 \cdot {{\log }_2}\left( {{2^{2015}}} \right) + 5 \cdot 2015}}{{20{{\log }_2}\left( {{2^{2015}}} \right)}} = \frac{{1995 \cdot 2015 + 5 \cdot 2015}}{{20 \cdot 2015}} = \frac{{2000}}{{20}} = \boxed{100}$