I trt to get Mathematica getting me the simple solutions for x and y from this system of two power equations:
$$\text{Eq. 1: } z=x^{m}y^{n}$$ $$\text{Eq. 2: }z=\tau x^{m-1}y^{1-n},$$
The simple solutions are
$$x=\left(z^{1-2n}\tau^{-n}\right)^{\frac{1}{n-2nm+m}}$$ $$y=\left(z\tau^{m}\right)^{\frac{1}{n-2nm+m}},$$
but I cannot make Mathematica find them. Entering in Mathematica
Solve[z == x^m y^n && z == \[Tau] x^(m - 1) y^(1 - n), {x, y}]
or, e.g.,
FullSimplify[Solve[z == x^m y^n && z == \[Tau] x^(m - 1) y^(1 - n), {x, y}]]
I get back
x -> E^((-Log[z] + 2 n Log[z] - n Log[\[Tau]])/(-m - n + 2 m n)),
y -> E^((-Log[z] + m Log[\[Tau]])/(-m - n + 2 m n))
that is,
$$x\to e^{\frac{-n \log (\tau )+2 n \log (z)-\log (z)}{2 m n-m-n}},y\to e^{\frac{m \log (\tau )-\log (z)}{2 m n-m-n}}.$$
Why does even FullSimplify (or Simplify) not give me back the simple solution?
Even after imposing the simple result, i.e. defining
x = (z^(1 - 2 n) \[Tau]^-n)^(1/(n - 2 n m + m)); y = (z\[Tau]^m)^(1/(n - 2 n m + m));
Mathematica doesn't tell confirm to me that the difference between the LHS and the RHS of Equation 1,
x^m y^n - z
is zero:
FullSimplify[x^m y^n - z]
yields
-z + ((z\[Tau]^m)^(1/(m + n - 2 m n)))^n ((z^(1 - 2 n) \[Tau]^-n)^(1/(m + n - 2 m n)))^m
This is easily seen to be zero, but it puzzles me that Mathematica doesn't tell me that.
TrigFactor
does the trick. $\endgroup$