# logarithms power equation

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.

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Note: You aren't trying to "prove" the inequality (that would mean, showing that the inequality is true for any value of $x$ for which it makes sense). Instead, it seems you are trying to solve the inequality (that is, find all the values of $x$ for which the inequality is true). – Arturo Magidin Oct 12 '11 at 20:35
@Nahum Litvin: If what you want to do is to solve the inequality, then you should (a) change the title to something like "logarithm power inequality" and (b) change "to prove the following" to "solve the following inequality" and (c) in the third line, change "equation" to "inequality." Conveniently, the changed question would be the one that Arturo Magidin answered. – André Nicolas Oct 13 '11 at 3:12

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 $$x^{\log_3x} = x^r = (3^r)^r.$$
If you are trying to "solve for $x$", then you want to have $3+2r\lt r^2$; since $r^2-2r-3=(r-3)(r+1)$, the values where $r^2-2r-3\gt 0$ are when $r\gt 3$ or when $r\lt -1$. That is, when $\log_3(x)\gt 3$ (which means $x\gt 3^3$); or when $\log_3(x)\lt -1$ (which means $x\lt 3^{-1}$). So the solution set for the inequality are all $x$ with $0\lt x \lt \frac{1}{3}$ ($x$ needs to be positive so we can take its logarithm base $3$), and also all $x$ with $x\gt 27$. That is, $(0,\frac{1}{3})\cup(27,\infty)$. You do not get $x=27$. Note that it is perfectly possible for $\log_3(x)$ to be negative. – Arturo Magidin Oct 12 '11 at 20:30