I have a system of 2 non-linear equations with two variables:
1) $e^n - e^m = 1$
2) $-\ln(e^{-m} ( e^{nr} + e^{m})) + nr + \ln ( e ^{-m} (e ^{m} + 1 ))=n $
, where $n$ and $m$ are the two variables and $r$ is a constant. Also $m≠n$ and $n>0$.
I am having trouble solving the system. Can anybody please help? Stuck for days now. Also how do we convert each equation in their respective linear form? PLEASE! Thank you.
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From the second equation, we have: $$\ln(\frac {e^m+1}{e^m+e^{nr}})=n(1-r)$$ So, $$\frac {e^n}{e^{nr}}=\frac {e^m+1}{e^m+e^{nr}}$$ Cross multiplying, $$e^{m+n}+e^{n(r+1)}=e^{nr+m}+e^{nr}$$ So, $$e^{nr}(e^n-1)=e^{m}(e^{nr}-e^{n})$$ So, from the first relation, we get: $$e^{n}=0$$ This means that $e^m=-1$, which is impossible, so the equations have no real solutions.
Edit: The only other possibility is that $e^{-m}=0$, and $n=0$, which I did not count because that is not a real solution either.
Your second relationship can be written
$$\operatorname{ln} \frac{e^m+1}{e^{nr}+e^m}=n-nr$$
otherwise said, taking the exponential of both sides:
$$\frac{e^m+1}{e^{nr}+e^m}=e^{n(1-r)}$$
Setting $u=e^m, v=e^n$, your system can be transformed into
$$\begin{cases}v-u&=&1\\ \frac{u+1}{v^r+u}&=&v^{1-r}\end{cases}$$
As you can "extract" $u=v-1$ from the first relationship, and plug its expression into the second one, you are left with :
$$\dfrac{v}{v^r+v-1}=v^{1-r}$$
Simplifying it by $v \ne 0$ (it is an exponential), we get:
$$v=0$$
which is contradictory.