$4^x+6^x=9^x$ $\implies$ $x \notin \mathbb Q$? Does there exist any rational number $x$ such that  $4^x+6^x=9^x$ ? 

Divide through by $4^x$ to get $1 + \left(\dfrac{3}{2}\right)^x = \left(\dfrac{9}{4}\right)^x$, i.e. $\left(\dfrac{3}{2}\right)^{2x} - \left(\dfrac{3}{2}\right)^x - 1 = 0$. 
Using the quadratic formula, we get $\left(\dfrac{3}{2}\right)^x = \dfrac{1\pm\sqrt{5}}{2}$. 
Since $\left(\dfrac{3}{2}\right)^x > 0$, we must have $\left(\dfrac{3}{2}\right)^x = \dfrac{1+\sqrt{5}}{2}$, i.e. $x = \log_{3/2}\left(\dfrac{1+\sqrt{5}}{2}\right)$. 
Since this is the only solution, you just need to show that this number is irrational. 
Suppose $x$ is rational, i.e. $x = \dfrac{a}{b}$ for some integers $a,b$ with $b > 0$. 
Then, $\left(\dfrac{3}{2}\right)^{a/b} = \dfrac{1+\sqrt{5}}{2}$, i.e. $\dfrac{3^a}{2^{b-a}} = (1+\sqrt{5})^b$. Clearly, $\dfrac{3^a}{2^{b-a}}$ is rational. 
However $(1+\sqrt{5})^b = c+d\sqrt{5}$ for some positive integers $c,d$ (Use the binomial theorem here). 
Thus, $(1+\sqrt{5})^b$ is irrational, a contradiction. Therefore, $x$ is irrational. 
