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so I have this inequality: Given integers $m, k\geq1$.

$$2^{m/k} > \frac{3}{2}$$

I'm interested in finding the smallest integer power $m$, as a function of $k$, that will make this inequality hold true.

I've made a few calculations and came up with $m$ as: $$m(k) = [ 6k/10 + 0.5 ] - [k/70]$$

I'm almost certain that this $m$, for any given $k$, will satisfy the inequality. I believe it is the smallest integer (I don't know how to prove it, a proof will be appreciated!), but I wonder if there's a way to find another m that's a lot simpler... or maybe a way to simplify the m I found so I can use it in power additions and other things.

Or in general, is there a way/technique for such problems on how to find the smallest integer power?

Thank you for your time!

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Given a positive integer $k$ you want to find the least positive integer $m$ such that $2^{m/k}>\frac{3}{2}$, or equivalently $$\frac{m}{k}>\frac{\log(\frac{3}{2})}{\log2}=\frac{\log3-\log2}{\log2}=\frac{\log3}{\log2}-1,$$ or equivalently $m>k\cdot\left(\frac{\log3}{\log2}-1\right)$. The latter is a constant; an online calculator tells me that $$\frac{\log3}{\log2}-1\approx0.5849625,$$ so $m=\lceil0.5849625\ldots\times k\rceil$.

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  • $\begingroup$ Thank you for your quick answer! So m should be [k*0.5849625 + 0.5] correct? $\endgroup$ – Buk Lau Dec 2 '18 at 22:12
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HINT

We have that since $\log$ function is strictly increasing

$$2^{m/k} > 3/2 \iff \log (2^{m/k}) > \log (3/2)$$

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