Given a fast and a slow computer, at what sizes does the fast computer running a slow algorithm beat the slow computer running a fast algorithm?
Consider your last inequality:

$\qquad \displaystyle n &lt; \frac{c_2y}{c_1x}\log(n)$

Now, $C\log n$ with $C = \frac{c_2y}{c_1x}$ is a concave function. Therefore, there are at most two intersections, and only one of them is interesting for you; that is, solve

$\qquad \displaystyle n = C\log(n)$

and find out which algorithm is faster in which interval by simply calculating the respective function values at some point in the respective interval.

It is true that solving this equality explicitly is hard/impossible. For some fixed $C$, computer algebra gives you an expression in a well-known function; interesting (real) intersection turns out to be at

$\qquad \displaystyle n = e^{-W_{-1}(-\frac{\ln 2}{C})}$

if $C \geq e \ln 2$, with $W_k$ the analytic continuation of the product log function. This function does not have a nice closed form, but it can be evaluated numerically for given $C$; for instance, you get $\approx 116,74$ for $C=17$.