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

Question

$3^{x^2} < 9^x$

According to my book the solution is $x>0 \land x < 2$ 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

$3^{x^2}-3^{x2}<0$

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?

Answer 1

Use the laws of exponents.

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

The rest follows trivially.

Answer 2

$3^{x^2}<9^x \iff$

$3^{x^2}<3^{2x} \iff$

$x^2<2x \iff$

$x(x-2)<0 \iff$

$0<x<2$.

Answer 3

You can use that $3^x$ is an increasing function. Hence $$3^{x^2}<3^{2x}\iff x^2<2x\iff x^2-2x=x(x-2)<0\iff 0<x<2.$$