2019/019) Solve for positive x $4^x+6^x = 9^x$

we have $4=2^2$, $6= 2 * 3$, $9=3^2$
So we get
$2^{2x}+( 2 * 3)^x = 3^{2x}$
divding by $2^{2x}$ we get
$1 + (\frac{3}{2})^x =   (\frac{3}{2})^{2x}$
Putting $(\frac{3}{2})^x = y$ we get
$1 + y = y^2$
so y = golden ratio $\phi = \frac{1+\sqrt{5}}{2}$
Giving
$(\frac{3}{2})^x = \frac{1+\sqrt{5}}{2}$
taking log on both sides we get
$x= \frac{\log \frac{1+\sqrt{5}}{2}}{\log \frac{3}{2}}$