As I know logarithm over the reals define only if base >0 and not equal 1. By this way, why Wolfram give an answer: (all values of x are solutions) for
solve log[-x,1]=0 over the reals
https://www.wolframalpha.com/input?i=solve+log%5B-x%2C1%5D%3D0+over+the+reals
I think we need to return to function domain. By the definition log[-1,1] - not defined.
It is the exponential function mapping n to x^n. Algebraically, it is a homomorphism of the additive integers into the multiplicative reals.
n
x^n
The exponential function is perfectly defined for negative basis and integer exponent. Take for instance (-1)^n.
(-1)^n
The trouble starts with negative basis and noninteger exponent.
This is a special case, since (-x)^0 = 1 makes sense and is true for all x. Other examples are Log[-x, -x] == 1 and Log[-x, x^2] == 2. What has no solution over the reals is for example the log of 2 in base -4.
(-x)^0 = 1
x
Log[-x, -x] == 1
Log[-x, x^2] == 2
-4
