Chebyshev used the divisibility of the middle binomial coefficient ${{2n}\choose{n}}$ to set upper and lower bounds on the number of primes in the form

$\frac{an}{\log(n)} > \pi(n) > \frac{bn}{\log(n)}$

for some constants $a$ and $b$. If $a$ is not too small and $b$ is not too large then the proofs can be elementary and quite short. See e.g. http://www.fen.bilkent.edu.tr/~franz/nt/cheb.pdf for proofs with $a = 6 \log(2)$ and $b = \log(2)/2$

There must then be a prime between $x$ and $nx$ if

$ \frac{ax}{\log(x)} < \frac{bnx}{\log(nx)} $

which is equivalent to

$a \log(nx) < b n \log(x)$

so for the $a/b = 12$ as above you get

$12 \log(n) < (n-12) \log(x) $

$ x > \exp\left({\frac{12 \log(n)}{n-12}}\right)$.

So, for any value of $n > 12$ this gives a lower bound for $x$ above which there is always a prime between $x$ and $xn$; for smaller $x$ it can be checked by hand.