# Smallest integer power for an inequality to hold

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?

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 $$m=\lceil0.5849625\ldots\times k\rceil$$.
We have that since $$\log$$ function is strictly increasing
$$2^{m/k} > 3/2 \iff \log (2^{m/k}) > \log (3/2)$$