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No output from Solve[ ]?

Could someone please look at this and tell me what's wrong? It's probably a simple mistake. I think this system of equations should return {{pd -> -3}, {pd -> 10}}. At least, that's what I get when I use a pencil and paper. But my input below doesn't return anything. Not even an error message.

POSTED BY: Jay Gourley
6 Replies

Check the syntax: Solve does not take Assumptions, you give them with the equation itself. What values of a,b,c did you use? With a = 2, b = 4, y = 2 I get no solution:

With[{a = 2, b = 4, y = 2}, 
 Solve[y == 2^c a (4 + 4 (a/b)^(3/c))^(-c/3), c, Reals]]
With[{a = 2, y = 2, b = 4},
 Plot[y - 2^c a (4 + 4 (a/b)^(3/c))^(-c/3), {c, -4, 4}]]

Reduce gives False:

Reduce[y == 2^c a (4 + 4 (a/b)^(3/c))^(-c/3) && a < y < b && a > 0 && 
  b > 0, c, Reals]

Anyway, with respect to c the equation is tascendental, very unlikely to have a symbolic solution with symbolic coefficients.

POSTED BY: Gianluca Gorni

The solution set is more complicated than that:

Reduce[(js + ps) 6 == w && js (pd + 5) == w && ps pd == w, 
  Reals] // LogicalExpand

I don't know the exact reason why Solve gives an empty set in that special case, but with slight changes in syntax we get meaningful answers:

eqs = (js + ps) 6 == w && js (pd + 5) == w && ps pd == w;
Solve[eqs, Reals]
Solve[eqs, {pd, w, ps, js}, Reals]
Solve[eqs]
POSTED BY: Gianluca Gorni
Posted 1 year ago

I have similar issue with other equation. I have no idea how to solve it. It works fine for predefined values of a,b,c, however I get empty list as output for exact solution.

Solve[y == 2^c a (4 + 4 (a/b)^(3/c))^(-c/3), c, Reals, 
 Assumptions -> a < y < b && a > 0 && b > 0]

Any idas how to deal with it? Thank you.

POSTED BY: Luka Roblek

(1) When you have a new question, it should be posted as such rather than appended to a different thread.

(2) I cannot imagine how one might construct an analytic solution to this equation.

POSTED BY: Daniel Lichtblau
Posted 1 year ago

Solution is only possible for a < y < b. For example with a = 2, b = 4, y = 3, the solution is c = -3.

An exact solution may not be possible. I was just asking if anyone knew of any alternative approach to solve the problem.

POSTED BY: Luka Roblek

Sorry, I chose a wrong value for y. My conjecture is that there is a single solution for all y between a and b, except when y == Sqrt[a b], which is the limit of the right-hand side as c goes to +Infinity and to -Infinity. But I doubt that there is a formula for the solution.

POSTED BY: Gianluca Gorni
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