<s>[INST] Real Solution of $4^x + 6^x = 9^x$. So here is the problem 

Solve $$4^x + 6^x = 9^x$$ for $x$.

I am trying to find its real solution. 
I was trying in this way!!
$$6^x\left( \frac {4^x}{6^x} + 1\right) = 9^x$$
$$\left({\frac 23}\right)^x +1=\left({\frac 32}\right)^x$$ but I'm stuck . . .
Will anyone help ?
 [/INST] $$4^x + 6^x = 9^x \implies \left(\frac{4}{9}\right)^x +\left(\frac{6}{9}\right)^x =1 \implies \left(\frac23\right)^{2x}+\left(\frac23\right)^{x}-1= 0 $$
Let $\left(\frac23\right)^x = y$
So, $$y^2+y-1 = 0 \implies y =\left(\frac23\right)^x = \frac{-1\pm \sqrt{5}}{2}$$
Considering the positive root, $$x = \frac{\ln y}{\ln \frac23} =\frac{\ln\left(\frac{-1+ \sqrt{5}}{2}\right)}{\ln \frac23} $$
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