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.
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]
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.
(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.
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.
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.
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.