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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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    @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.2011-10-13

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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?

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    so delete it sorry2011-10-12