How to deal with three equations. We had a lesson in chemistry when we learned this equation  $R=K(A)^n*(B)^m$.
They give you a table with the value of R,a and b in three different situations(different value) to get the rest (k,n and m)
I can't find a way to make one equation once for n and m and other one for k so
I can find its value with only a,b and r



for example

   A= B= R=



2          3        216
4          2        256
3          4        1152


so what is n, and m and k
by the equation I will find that
n=2
m=3
k=2
So you are given that
$$\begin{align} K\cdot 2^n\cdot 3^m&amp;=\hphantom{0}216=2^33^3\\
K\cdot 4^n\cdot2^m&amp;=\hphantom{0}256=2^8\\
K\cdot 3^n\cdot4^m&amp;=1152=2^73^2\end{align}$$
Assuming for simplicity that $K=2^u3^v$ for some $u,v$ and nothing messy happens, we obtain the following equations by considereing the exponents of $2$ and $3$ in the given equations:
$$u+n=3\qquad v+m=3 $$ 
$$u+2n+m=8\qquad v=0 $$ 
$$u+2m=7\qquad v+n=2 $$
which readily leads to $v=0,n=2,m=3,u=1$. The latter system of six equations is overdetermined, so we need to check all equations and are somewhat lucky that this simple trick works out (this is not guaranteed with more "random" numbers as input).

(The right method for the general case is taking logarithms and solving three linear equations in three unknowns $n,m,\ln K$, as Yves Daoust suggests)
Take the logarithm and set $k=\ln K$:
$$n\ln2+m\ln3+k=\ln216,\\n\ln4+m\ln2+k=\ln256,\\n\ln3+m\ln4+k=\ln1152.$$
Then by pairwise subtraction
$$n\ln2+m(\ln2-\ln3)=5\ln2-3\ln3,\\n(\ln3-2\ln2)+m\ln2=2\ln3-\ln2.
$$

By inspection (or by Cramer), $n=2$ and $m=3$. $k$ follows.