Exponential Function: $4^x + 6^x = 9^x$

Dividing both sides by 6^x, you get
4^x/6^x+1=9^x/6^x
Simplifying the fractions you get
2^x/3^x+1=3^x/2^x
That can be written as
(2/3)^x+1=(3/2)^x
Now, if the number you want is (2/3)^x=k, then (3/2)^x=1/k ,
which means the equation can be written as
k+1=1/k
Multiplying both sides times k,
k²+k=1
Solving that quadratic equation gives you two real solutions for k.

I am not used anymore to calculate this type of equation, the result of relying too much in SHIFT + SOLVE of Casio. I did not say doing Shift + Solve is bad, it is actually highly recommended in my line of work, hehehe. Anyway, allow me to solve this, and I am actually surprised that solving for x is more complex than solving for (2/3)^x. Here is my take based on the suggestion of KMST.
$4^x + 6^x = 9^x$

$\dfrac{4^x}{6^x} + \dfrac{6^x}{6^x} = \dfrac{9^x}{6^x}$

$\left( \dfrac{4}{6} \right)^x + 1 = \left( \dfrac{9}{6} \right)^x$

$\left( \dfrac{2}{3} \right)^x + 1 = \left( \dfrac{3}{2} \right)^x$

$\left( \dfrac{2}{3} \right)^x + 1 = \dfrac{1}{\left( \dfrac{2}{3} \right)^x}$

$\left( \dfrac{2}{3} \right)^{2x} + \left( \dfrac{2}{3} \right)^x = 1$

$\left( \dfrac{2}{3} \right)^{2x} + \left( \dfrac{2}{3} \right)^x - 1 = 0$
 

By Quadratic Equation
$\left( \dfrac{2}{3} \right)^x = \dfrac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)}$

$\left( \dfrac{2}{3} \right)^x = \dfrac{-1 \pm \sqrt{5}}{2}$
 

Hence,
$\left( \dfrac{2}{3} \right)^x = \dfrac{\sqrt{5} - 1}{2}$   ←   this is my answer.

Please be gentle with me, hehehe. Although I hope I was able to consider everyhting, maybe I miss simething. As I've said, I am no longer used to this type of approach. A decimal number from calculator is more than acceptable for me.