Find value or $W=\sqrt[4x+4]{3^{x}\left ( \sqrt{5}-1 \right )}$ according to condition $9^{x}=6^x +4^x$ If     $$9^{x}=6^x +4^x$$
Find the value of:
$$W=\sqrt[4x+4]{3^{x}\left ( \sqrt{5}-1 \right )}$$
Solving the equation arrives at:
$x = \frac{\log\left ( \sqrt{5}-1 \right ) - \log(2)}{\log(2) - \log(3)}$
But W yields a giant result,
Is there any algebraic manipulation that I do not see

Note that 
$$1=\left(\frac23\right)^x+\left(\frac49\right)^x$$
Let $a=\left(\frac23\right)^x$
$$a^2+a-1=0$$
Solving gives
$$a=\left(\frac{\sqrt5 -1}{2}\right)$$
$$\left(\frac23\right)^x=\frac{\sqrt5 -1}{2}$$
$$\left(\frac32\right)^x=\frac{2}{\sqrt5-1}$$
$$3^x=\frac{2^{x+1}}{\sqrt5-1}$$
Substitute for this in question
$$\left(\left(\frac{2^{x+1}}{\sqrt5-1}\right)(\sqrt5-1)\right)^{\frac{1}{4(x+1)}}=2^{\frac{x+1}{4(x+1)}}=2^{1/4}$$
