Solve for $x$, $3^{x^2} \lt 9^x$ $3^{x^2} &lt; 9^x$

According to my book the solution is $x&gt;0 \land x &lt; 2$ so I assume this is a quadratic equation.

Therefore, I need to have $variable^2+variable+ \not variable$.

But I don't know what variable  is in this case.

I have:

$3^{x^2}$

$9^x = (3^2)^x = 3^{2x}$

and so

$3^{x^2}-3^{x2}&lt;0$

So I think that variable should be $3^x$, so that in:
$$ax^2 + bx +c,$$
$$a = 1$$
$$b = ???$$
$$c = 0$$

But I don't know what b should be in this case... 

Am I thinking right? How do I solve this?
You can use that $3^x$ is an increasing function. Hence 
$$3^{x^2}&lt;3^{2x}\iff x^2&lt;2x\iff x^2-2x=x(x-2)&lt;0\iff 0&lt;x&lt;2.$$
Use the laws of exponents.

That is, the corresponding equation is  $3^{(x^2)}=3^{(2x)}$; $x^2=2x$. 

The rest follows trivially.