# How to solve the given equation?

I would like to solve the following equation,

Solve[(c + 1)*(u - 1) + c*(k + 1)*(u - 1)*(u^(1/k) - 1) -
c*k*u*(u^(1/k) - 1) == 0, u]


c and k are constants. But I am not getting the solution. Instead of solve, I tried Reduce and NSolve. Is there any other method to solve this equation.

• Reduce[(c + 1)*(u - 1) + c*(k + 1)*(u - 1)*(u^(1/k) - 1) - c*k*u*(u^(1/k) - 1) == 0] derives a solution u==1. Commented Mar 15, 2022 at 8:25

First, the command

Reduce[(c + 1)*(u - 1) + c*(k + 1)*(u - 1)*(u^(1/k) - 1) -     c*k*u*(u^(1/k) - 1) == 0]


(-k - u^(1 + 1/k) + u^(1/k) + k u^(1/k) != 0 && c == (-1 + u)/(-k - u^(1 + 1/k) + u^(1/k) + k u^(1/k))) || u == 1

derives a solution u==1. Second, one can symbolically solve  c == (-1 + u)/(-k - u^(1 + 1/k) + u^(1/k) + k u^(1/k))) for natural values k==1,k==2,k==3, and k==4. E.g.

k = 2; Solve[c == (-1 + u)/(-k - u^(1 + 1/k) + u^(1/k) + k u^(1/k)), u]


{{u -> (1 + 5 c^2 - (1 + c) Sqrt[1 - 2 c + 9 c^2])/(2 c^2)}, {u -> ( 1 + 5 c^2 + (1 + c) Sqrt[1 - 2 c + 9 c^2])/(2 c^2)}}

Third, one can numerically solve that equation for some real values of k and c. E.g.

c = 1/E; k = Pi; Plot[c - (-1 + u)/(-k - u^(1 + 1/k) + u^(1/k) + k u^(1/k)), {u, 0, 50}]


suggests a possible root near the origin, so

FindRoot[c == (-1 + u)/(-k - u^(1 + 1/k) + u^(1/k) + k u^(1/k)), {u,0}]


{u->0.000785876}