- From NRICH,

find all real solutions to

Extension: What if x is permitted to be a complex number?

- MathHelp replies,

### First tryEdit

*a*^{b}=1 when *a*=1 or *b*=0 (and *a* is nonzero), so the first try at a solution is to solve

and

and take the union of the results.
The first equation gives us and . The second equation gives us 1 and −1. But then we need to throw out because that solution gives us 0^{0}, which is undefined. So the results of the first try are:

### Second tryEdit

Noting that (-1)^{2}=1 and that for all integers, *k*, there are certainly other solutions to *a*^{b}=1. For all I know at this point, there may be real solutions in x to the pair of equations

and

where *a* and *b* are complex solutions to *a ^{b}*=1.

So I would like to start by characterizing solutions to *a ^{b}*=1. First, the equation is solved whenever

*a*=1, regardless of

*b*. If

*a*is not equal to 1, then,

*a*= e^{b}^{b ln(a)}=

and so the complete solution to *a ^{b}*=1 is

The second solution can also be expressed as

^{[improvement needed: You can help]}