Golden Problem.

Let $4^{x} + 6^{x} = 9^{x}$

Dividing both sides by $4^{x} \implies 1 + (\dfrac{3}{2})^{x} = ((\dfrac{3}{2})^{x})^{2}$

Let $u = (\dfrac{3}{2})^{x} > 0 \implies 1 + u = u^{2} \implies u^{2} - u - 1 = 0 \implies$ $u = \dfrac{1 + \sqrt{5}}{2} \implies \ln(u) = \ln(\dfrac{1 + \sqrt{5}}{2}) = x\ln(\dfrac{3}{2}) \implies$ $x = \dfrac{\ln(\dfrac{1 + \sqrt{5}}{2})}{\ln(\dfrac{3}{2})} = \dfrac{\ln(\dfrac{a + \sqrt{b}}{c})}{\ln(\dfrac{d}{c})}$

$\implies a + b + c + d = \boxed{11}$ .