Task: Let $x,y≠0$ with $x+y-2xy=0$, such that $4^x=9^y=a$. Then, what is the value of $a$?
My attempts.
Let, $x=\log_2m,\, y=\log_3n$ then using $\frac 1x+\frac 1y=2$ gives,
$$\begin{align}m^2=n^2&=a\\ \log_{m^2}2^2+\log_{n^2}3^2=\log_a{36}&=2\\ a&=6.\end{align}$$
Teacher's original solution:
$$\frac 1x+\frac 1y=2$$
$$4=a^{\frac 1x}, 9=a^{\frac 1y}$$
$$36=a^{\frac 1x+\frac 1y}=a^2\\ a=6.$$
Backround:
Sometimes I try to help neighbor children in their lessons, which they are stuck in, only at the level of precalculus mathematics that I know. I learned that the teacher rejected the solution I taught and said it was an unnecessary, long and fuzzy solution.
I accept that the teacher's solution is very nice and more elegant. I wanted to ask the question here for to get the mathematical opinion.
Is everything okay with my attempts? Maybe I missed some important points?