# How can I drop the "belongs to integers" part from Reduce output?

I have this Mathematica line:

Reduce[1/(E^0)^x + 1/(E^1)^x + 1/(E^3)^x == 0, x]


that gives the output:

C[1] ∈
Integers && (x ==
I π + 2 I π C[1] + Log[-Root[1 + #1^2 + #1^3 &, 1]] ||
x == 2 I π C[1] + Log[Root[1 + #1^2 + #1^3 &, 2]] ||
x == 2 I π C[1] + Log[Root[1 + #1^2 + #1^3 &, 3]])


when copy pasted into the question box at Mathematica Stackexchange.

I would like to drop the part:

C[1] ∈
Integers &&


from the beginning of the output above so that I can apply the ToRules command like this:

{ToRules[(x ==
I π + 2 I π C[1] + Log[-Root[1 + #1^2 + #1^3 &, 1]] ||
x == 2 I π C[1] + Log[Root[1 + #1^2 + #1^3 &, 2]] ||
x == 2 I π C[1] + Log[Root[1 + #1^2 + #1^3 &, 3]])]}


How can I get rid of the C[1]∈Integers && in the output above from the program involving Reduce?

• Why the down vote? Nov 20, 2014 at 12:54
• What is log3?
– Öskå
Nov 20, 2014 at 12:54
• log3 is just a variable. Could have been "y" also. Nov 20, 2014 at 12:55
• Why not List[ToRules@(Reduce[1/(E^0)^x + 1/(E^1)^x + 1/(E^3)^x == 0, x] /. C[1] -> 0)]?
– Öskå
Nov 20, 2014 at 13:00
• In this particular problem, the output from Reduce has head And and since Last works with any head if you do, say, sol = Reduce[1/(E^0)^x + 1/(E^1)^x + 1/(E^3)^x == 0, x]];, then Last@sol will give you what you want.
– gpap
Nov 20, 2014 at 13:23

I usually use Simplify with assumptions:

Simplify[
Reduce[1/(E^0)^x + 1/(E^1)^x + 1/(E^3)^x == 0, x],
C[1] ∈ Integers]
(*
x == I π (1 + 2 C[1]) + Log[-Root[1 + #1^2 + #1^3 &, 1]] ||
x == 2 I π C[1] + Log[Root[1 + #1^2 + #1^3 &, 2]] ||
x == 2 I π C[1] + Log[Root[1 + #1^2 + #1^3 &, 3]]
*)


That way I'm fairly confident the transformations are mathematically sound (as long as I keep to the assumptions).