# How to solve a symbolic equation with symbols in powers

I am trying to solve

$\dfrac{\rho\ ((1-b)\ \lambda _f\ \lambda _s\ P _A)^{\rho }}{b-1}+\dfrac{\rho \left(\dfrac{b\ \text{ps}\ x}{1-\text{ps}}\right)^{\rho }}{b}=0$

in terms of $b$,

I entered

   Solve[((((b ps x)/(1 - ps))^ρ ρ)/
b + ((-(-1 + b) PA λf λs)^ρ ρ)/(-1 + b)) == 0, b] // Simplify


but Solve function results to this message

!Solve::nsmet: This system cannot be solved with the methods available to Solve. >>{1]

How can I solve this equation?

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Please post the actual Mathematica code you have used. You can use the "edit" link below your question to update it. –  Sjoerd C. de Vries Sep 22 '12 at 7:01
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If you simplify by hand a bit your equation (removing irrelevant bits relative to b): eqn = (b ((ps x)/(1 - ps))^[Rho] + (1 - b) (P[CapitalAlpha] [Lambda]f [Lambda]s)^[Rho]) == 0; Solve[eqn, b] works; –  chris Sep 22 '12 at 7:30
Thank you all very much, I am new to this kind of site, I registered today because I am having problems with mathematica, your advices were really helpful. I will try again –  Jones Sep 22 '12 at 7:56
@Chris, you could give that as an answer... –  Sjoerd C. de Vries Sep 22 '12 at 23:22

Your equation needs such a massive amounts of assumptions that I can imagine Mathematica can't find the solution. Here is my rough manual derivation in Mathematica's TraditionalForm side-stepping many of the intricacies that will prevent Mathematica from solving this (for instance, I assumed $\rho\neq0$). A numerical test in Mathematica follows the derivation.

Table[((((b ps x)/(1 - ps))^\[Rho] \[Rho])/
b + ((-(-1 + b) PA \[Lambda]f \[Lambda]s)^\[Rho] \[Rho])/(-1 + b))
/. b -> 1/(1 + ((\[Lambda]f*\[Lambda]s*PA (1 - ps))/(ps*x))^(\[Rho]/(1 - \[Rho])))
/. {ps -> Random[],x -> Random[], \[Rho] -> Random[],
\[Lambda]f -> Random[], \[Lambda]s -> Random[], PA -> Random[]},
{100}
] // Chop // Quiet


{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, Indeterminate, 0, Indeterminate, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, Indeterminate, 0, 0, -6.9383157*10^-9, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1.196618005*10^-8, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1.205482381*10^-9, Indeterminate, 0, Indeterminate, 0, 0, 0, 0, Indeterminate, 0, Indeterminate, 0, 0, 0, Indeterminate, 0, 0, 0}

The Indeterminates stem from division by (near) zero, where the Random function generates a pole.

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You also assumed $\rho\neq1$ obviously. –  chris Sep 23 '12 at 14:18
@chris Yeah, I wrote "for instance, I assumed $\rho\neq0$" as an obvious example of what to exclude. There are more if you like: \[Rho] > 0 && b > 0 && \[Rho] != 1 && b != 0 && b != 1 && b > 0 && ps != 1 && x != 1 && ps != 0 && ps > 0 && x > 0 && PA > 0 && \[Lambda]f > 0 && \[Lambda]s > 0. I used these in Assuming, but that didn't help much. –  Sjoerd C. de Vries Sep 23 '12 at 14:22
Quite a few indeed :-) –  chris Sep 23 '12 at 14:23