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 ?
Hint:
If $\left(\dfrac32\right)^x=y>0$ for real $x$
$$y=1+\dfrac1y\iff y^2-y-1=0$$
$$y=\dfrac{1\pm\sqrt5}2$$
So, $\left(\dfrac32\right)^x=\dfrac{1+\sqrt5}2$
$4^x + 6^x = 9^x$ Divide through by $4^x$ $1 + (3/2)^x = (3/2)^{2*x}$
Now it's easier to deal with $x = \log[{3/2}]((1+\sqrt(5))/2)$
$$4^x + 6^x = 9^x \implies \left(\frac{4}{9}\right)^x +\left(\frac{6}{9}\right)^x =1 \implies \left(\frac23\right)^{2x}+\left(\frac23\right)^{x}-1= 0 $$
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, $$x = \frac{\ln y}{\ln \frac23} =\frac{\ln\left(\frac{-1+ \sqrt{5}}{2}\right)}{\ln \frac23} $$
