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 not exponent function, but power function. Let return to definition of logarithm function over the reals and domain of this function.
The problem is that exponential function is not defined for negative basis even for zero power over the reals
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
