TITLE: logarithms power equation
QUESTION [6 upvotes]: I got a home work question to solve the following:
$$ 27x^2 < x^{\log_3x} $$
can any one please explain how to solve this type of equation?
I have no idea what to do or what to search for.

REPLY [6 votes]: If $r=\log_3x$, then $3^r = x$. 
Since $27=3^3$, then you can rewrite the left hand side as
$$27x^2 = 3^3(3^r)^2 = 3^3\times 3^{2r} = 3^{3+2r}.$$
On the other hand, the right hand side would be
$$x^{\log_3x} = x^r = (3^r)^r.$$
Can you take it from here?