How can I transform exponential equation?

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Konrad Dadej
Konrad Dadej el 15 de Oct. de 2017
Comentada: Konrad Dadej el 6 de Nov. de 2017
I try to transform following exponential equation. I would like to obtain an analytical equation for x:
>> syms P A B x y z n
>> eqn1= P== A*((x+y)^n)+B*(x+z)
eqn1 =
P == A*(x + y)^n + B*(x + z)
>> solve (eqn1,x)
Warning: Unable to find explicit solution. For options, see
help.
> In solve (line 317)
ans =
Empty sym: 0-by-1
Any ideas why MATLAB can not solve above code?

Respuestas (1)

John D'Errico
John D'Errico el 15 de Oct. de 2017
Editada: John D'Errico el 15 de Oct. de 2017
Is there some reason why you think an analytical solution exists?
In fact, there is no analytical solution. Wanting magic to happen is not sufficient. How are your skills with a magic wand?
For example, suppose that n is greater than 4? There is a rather famous theorem showing that a polynomial equation of order greater than or equal to 5 has no analytical solution.
https://en.wikipedia.org/wiki/Abel–Ruffini_theorem
But in your case, n is completely unknown. n is a variable. So you have essentially a polynomial equation with non-constant coefficients, of completely unknown degree.
Only a crystal ball will suffice for you here.
  2 comentarios
Walter Roberson
Walter Roberson el 16 de Oct. de 2017
Yup. Let A = 1, B = 2, n = 5, y = 0, z = 1, P = 0, so 0 = 1*(x + 0)^5 + 2*(x+1), so x^5 + 2*x + 2 = 0 . There are no closed form solutions to that.
Konrad Dadej
Konrad Dadej el 6 de Nov. de 2017
Thank you for answer - it is clear.
However, I wonder if the following assumptions would make it possible to solve? P>0 A>0 B>0 x>0 0<n<1 -1<y<1 -1<z<1 Is anybody can try to solve it?

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