Let f : R $\to$ R satisfy $f(x + y) = {2^x}f(y) + {4^y}f(x)$, $\forall$x, y $\in$ R. If f(2) = 3, then $14.\,{{f'(4)} \over {f'(2)}}$ is equal to ____________.

<p>$\because$ $f(x + y) = {2^x}f(y) + {4^y}f(x)$ ....... (1)</p> <p>Now, $f(y + x){2^y}f(x) + {4^x}f(y)$ ...... (2)</p> <p>$\therefore$ ${2^x}f(y) + {4^y}f(x) = {2^y}f(x) + {4^x}f(y)$</p> <p>$({4^y} - {2^y})f(x) = ({4^x} - {2^x})f(y)$</p> <p>${{f(x)} \over {{4^x} - {2^x}}} = {{f(y)} \over {{4^y} - {2^y}}} = k$</p> <p>$\therefore$ $f(x) = k({4^x} - {2^x})$</p> <p>$\because$ $f(2) = 3$ then $k = {1 \over 4}$</p> <p>$\therefore$ $f(x) = {{{4^x} - {2^x}} \over 4}$</p> <p>$\therefore$ $f'(x) = {{{4^x}\ln 4 - {2^x}\ln 2} \over 4}$</p> <p>$f'(x) = {{({{2.4}^x} - {2^x})ln2} \over 4}$</p> <p>$\therefore$ ${{f'(4)} \over {f'(2)}} = {{2.256 - 16} \over {2.16 - 4}}$</p> <p>$\therefore$ $14{{f'(4)} \over {f'(2)}} = 248$</p>