You can get a solution from Solve
. Your equation can be rewritten
(1 + 2 r)^(1 + n) == (p + 2 s)/(2 b + p)
Making what I hope you can accept as a reasonable assumption about 2 b + p
, I evaluated
Assuming[n ∈ Integers && 2 b + p != 0,
Solve[(1 + 2 r)^(1 + n) == (p + 2 s)/(2 b + p), r]]
Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. >>
{{r -> 1/2 (-1 + (p/(2 b + p) + (2 s)/(2 b + p))^(1/(1 + n)))}}
In this case, Reduce
won't do any good, but perhaps you can live without knowing if the solution found by Solve
may not be complete. I think you need only note that the formal solution fails for n = -1
.
Assuming
only works with functions that take anAssumptions
option, whichReduce
doesn't. You can append&& Element[n, Integers]
to the equation instead. $\endgroup$