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I'm really struggling with this inequality. It seems I should solve it using logarithms but have no idea how to it specifically. Can you help me with that? $(2^x)*(3^{1/x})>6$

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closed as off-topic by Shailesh, Paramanand Singh, Brandon Carter, user284331, Dylan Mar 14 '18 at 21:43

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HINT

Take log both side and observe that log is strictly increasing, then

$$\iff x\log 2 + \frac1x \log 3 > \log 6$$

Can you proceed from here?

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  • $\begingroup$ That's a really good answer and I'm experimenting with it but actually I am still not able to proceed. Can you help a bit more if not difficult? $\endgroup$ – Michael Kaprenkov Mar 14 '18 at 9:35
  • $\begingroup$ @MichaelKaprenkov For the first step given by the hint please note that the fact that log is strictly increasing is a foundamental fact (if it was strictly decreasing the ineguality reverses). From here what we can do is multiply by $x\neq 0$ and obtain a quadratic equation but, note again, since we are dealing with inequality we must separate 2 cases x>0 and x<0 (x=0 is not admitted by the original equation). Can you proceed? $\endgroup$ – gimusi Mar 14 '18 at 9:39

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