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How could you solve the following system ?

$$x^y = z+15 \tag{1}$$ $$z^y = 20x\tag{2}$$ $$zy = 25 - x\tag{3}$$

Combining $(1)$ and $(2)$ yields a 2nd degree equation for $y$, which can then be expressed with $z$:

$$ln(z) * y^2 - ln(20) * y- ln(z+15) = 0\tag{4}$$ $$y = \frac{ln(20) \pm \sqrt{ln(20)^2 + 4*ln(z+15)*ln(z)}}{2*ln(z)} \tag{5}$$

Combining $(3)$ and $(5)$ easily allows to express $x$ with $z$ :

$$x = -z * \frac{ln( 20 ) \pm \sqrt{ln(20)^2 + 4*ln(z+15)*ln(z)}}{2*ln(z) } + 25 \tag{6}$$

But then I'm at a loss when I try to combine these two to find a value for $z$. Is it even possible? If not, how can one see it?

I do know $x=5,y=2,z=10$ are solutions but I want to understand, if possible, how to solve it, without numerical analysis obviously.

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Another real solution is approximately $x = 1.145619520$, $y = 20.47045460$, $z = 1.165307803$. I don't know if this can be expressed in "closed form".

EDIT: You can eliminate two variables to get a single equation in one.

Taking natural log of both sides of equations 1 and 2,

$$ \eqalign{y \ln(x) &= \ln(z+15) \cr y \ln(z) &= \ln(20) + \ln(x)\cr}$$ and then eliminate $\ln(x)$, so $$ y^2 \ln(z) - y \ln(20) = \ln(z+15) $$ Solve this quadratic for $y$: $$ y = \frac{\ln(20) \pm \sqrt{\ln(20)^2 + 4 \ln(z) \ln(z+15)}}{2 \ln(z)} \tag 4$$ It seems only the $+$ case produces real solutions. On the other hand, since $x = 25 - zy$ from equation 3, $$ y \ln(25 - z y) = \ln(z+15) $$ and we can substitute equation 4 in here to get a rather complicated equation for $z$.

A graph of the difference between the sides of that equation (for the $+$ case) looks like this:

enter image description here

There appear to be two real solutions, which are $z=10$ (which you mentioned) and $z = 1.165307803$.

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  • $\begingroup$ Thanks but I'm not really looking for solutions, but more for a way to solve it (or a way to see it's not possible). $\endgroup$
    – Bastien
    Commented Apr 19, 2020 at 20:17

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