$4^x+6^x=9^x$ $\implies$ $x \notin \mathbb Q$? Does there exist any rational number $x$ such that  $4^x+6^x=9^x$ ? 
 A: 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. 
A: Let $u=(3/2)^x$, then
$$4^x+6^x=9^x\implies 1+u=u^2$$
Therefore 
$$u=\frac{1+\sqrt{5}}{2}$$
$$x=\frac{\ln\left(\frac{1+\sqrt{5}}{2}\right)}{\ln(3/2)}$$
Please refer to the first solution by JimmyK4542 for proving $x$ is irrational.
A: The equation can be rewritten as 
$$1=\left(1+\left(\frac{2}{3}\right)^x\right)\left(\left(\frac{3}{2}\right)^x-1\right).$$ Using $t=\left(\frac{2}{3}\right)^x$ we get a quadratic
$t^2+t-1=0$ which solves to 
$$t=\frac{-1 \pm \sqrt{5}}{2}.$$
For obvious reasons we disregard the negative value and let $x=a/b$ with $a,b \in \mathbb{Z}$ with $b \neq 0$.
This gives
$$\left(\frac{2}{3}\right)^a= \left[\frac{\sqrt{5}-1}{2}\right]^b$$
The RHS can be shown to be irrational whereas the LHS is rational, so $x \not\in \mathbb{Q}$.
