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.
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.
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?