Solving an equaiton which includes $log$ as both base and exponent

Question

Q: If $$9x = x^{\log_3x}$$ then what is $x$ ?

I can't solve it. I have tried to use identities in my book but i think they are useless for this question. I need a hint

Answer 1

It is $$\log_3(9x)=(\log_3x)^2$$

Can you proceed from here?

Answer 2

Hint: Try taking logarithm to the base $3$ on both sides. You'll get a quadratic in $\log_3x$.

Answer 3

Take $\log_3$ of both sides. Then $$ \log_3{9} + \log_3{x} = (\log_3{x}) (\log_3{x}), $$ using $\log{ab}=\log{a}+\log{b}$ and $\log{a^b} = b\log{a}$. This is a quadratic equation for $\log_3{x}$. Solve that and exponentiate $3$ with the answer to find $x$.