# A question about the Solve function

I want to solve $Z$ in the following equation: $$(1 + \rho Z)^\beta (1 + 2 \rho Z)^{1-\beta} < 2^{3R/2} \tag{1},$$ where $\rho > 0$, $Z \geq 0$, $\beta \in (0, 1)$, and $R > 0$. Then I input the following code into Mathematica 11:

Solve[Z >= 0 && (1 + \[Rho] Z)^\[Beta] (1 + 2 \[Rho] Z)^(1 - \[Beta]) < 2^(3 R/2), Z]


After a long time (about 2 hours), the calculation aborted automatically and return no output. Does anyone know where the problem is ? Thank you very much in advance.

• Is $\beta$ integer? – yarchik Sep 7 '16 at 7:32
• @yarchilk No, $\beta \in (0, 1)$. – Wei-Cheng Liu Sep 7 '16 at 7:42
• And $1+\rho Z>0$ and $1+2\rho Z>0$ is fulfilled or you want to get it automatically ? By the way, by default all variables are assumed to be complex. I guess $R$ is real ? – yarchik Sep 7 '16 at 7:49
• @yarchik Yes, $R > 0$, $\rho > 0$, and $Z \geq 0$. – Wei-Cheng Liu Sep 7 '16 at 7:50
• Thank you, may be you can update your question little bit. Explain that the variables are real and positive, clarify what would you expect as the result. By this you greatly increase your chances that someone will provide a useful answer. – yarchik Sep 7 '16 at 7:55

The best you can do is to Solve for a particular β. First get an idea about your functions.

Manipulate[ Plot3D[{(1 + ρ Z)^β (1 + 2 ρ Z)^(1 - β), 2^(3 R/2)},
{ρ, 0, 1}, {Z, 0, 3}], {β, 0, 1}, {R, 0, 2}] You are looking for the yellow surface below the blue surface. You can have the solution for the arc for a particular β as,

Block[{β = 1/2}, Solve[(1 + ρ Z)^β (1 + 2 ρ Z)^(1 - β) == 2^(3 R/2), Z]]


It will give you 2 solutions. However, if you choose β = 0.13 there would be 100 solutions (because you are dealing with a polynomial). The can filter the solutions within $Z>0$ (or the Reals) only when you specify the other parameters. Once you have the arc, you know your region of solutions.