Interesting System of Equations to solve.

Question

I thought about this system of equations that I am interested in finding the solution to. I tried to solve it but wasn't able too since it isn't exactly easy to solve. I was wondering the best way to go about solving it without using brute force in a computer program. Is it even possible to come up with a good solution?

z has to be equal to naught, and it is not possible to use only integers.

$ \begin{cases} z*e*r*o = 0 \\ o*n*e = 1 \\ t*w*o = 2 \\ t*h*r*e*e = 3 \\ f*o*u*r = 4 \\ f*i*v*e = 5 \\ s*i*x = 6 \\ s*e*v*e*n = 7 \\ e*i*g*h*t = 8 \\ n*i*n*e = 9 \\ t*e*n = 10 \\ \end{cases} $

Answer 1

Note that there are eleven equations, but fifteen variables ($z$, $e$, $r$, $o$, $n$, $t$, $w$, $h$, $f$, $u$, $i$, $v$, $s$, $x$, and $g$). So there won't be a unique solution - at best, you could hope to find eleven of them in terms of the other four.

But taking the logarithm of each equation will make this doable: for example, $\ln{o * n * e} = \ln 1$ tells you that $\ln{o} + \ln{n} + \ln{e} = 0$. Then you can solve the system of linear equations for the logs of the variables; once you've done that, undo the logarithm to get the value of the variables.

Answer 2

The solutions can be expressed as $$ \eqalign{f&={\frac {144\,{e}^{2}v}{5\,{r}^{2}{u}^{2}}},g={\frac {384\,{ v}^{2}{e}^{4}}{25\,r{u}^{2}}},h={\frac {54\,v}{25\,{r}^{2}u}},i={ \frac {25\,{r}^{2}{u}^{2}}{144\,{e}^{3}{v}^{2}}},n={\frac {36\,ev}{5\, ru}},o={\frac {5\,ru}{36\,{e}^{2}v}},\cr s&={\frac {35\,ru}{36\,{e}^{3}{v}^ {2}}},t={\frac {25\,ru}{18\,{e}^{2}v}},w={\frac {1296\,{v}^{2}{e}^{4} }{125\,{r}^{2}{u}^{2}}},x={\frac {31104\,{e}^{6}{v}^{4}}{875\,{r}^{3}{ u}^{3}}},z=0} $$ with $e,r,u,v$ arbitrary nonzero numbers.