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? $(2^x)*(3^{1/x})>6$


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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shailesh, Paramanand Singh, Brandon Carter, user284331, Dylan
If this question can be reworded to fit the rules in the help center, please edit the question.



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?

  • $\begingroup$ 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

Not the answer you're looking for? Browse other questions tagged or ask your own question.