Is there an imaginary Exponent with $a^x=-b$ while a,b>0

Question

Is there an imaginary Exponent $x$ with $a^x=-b$ while a,b>0 ?

An where could something like this be used?

Answer 1

The way to define complex exponentiation is through the complex exponential map $z\mapsto e^z$. If you dont know what the complex exponential map is, it is defined by $e^{a+bi}:=e^ae^{ib}=e^a(\cos b +i\sin b)$, where the last equality used the well known Euler's identity.

Now, complex exponentiation is defined similarly to real exponentiation ($a^x:=e^{x\log a}$). That is, $a^z:=e^{z\log a}$ (this makes sense cause $a$ is a real positive number. It can be defined this for complex $a$ and we'd need to define a complex logarithm, but we dont need them here). It is clear then that the function $f(z)=a^z$, which is defined for every $z\in\mathbb{C}$, sends $\mathbb{R}$ to $(0,+\infty)$. Your question then can be stated as $$f(\mathbb{C}\setminus\mathbb{R})\cap(-\infty,0)\neq\phi?$$ The answer is yes: in order of $f(z)$ to be real and negative it is enough and sufficient that the imaginary part of $z$ lies on the set $\pi+2\pi\mathbb{Z}$, i.e. that (if $z=x+iy$ then) $y=\pi+2k\pi$ for some $k\in\mathbb{Z}$. This follows directly from the definition of $a^z$ given above. So every complex number of the form $x+i(\pi+2k\pi)$ (where $x$ is real and $k$ integer) is an exponent satisfaying the condition in the first question.

I don't think this is something with self name in terms of applicability. As mentioned, this is a particular case of the operation of complex exponentiation so, although one might argue that it is used implicitly, it does not tell anything deep; it is just an inmediate consequence of the definition of complex exponentiation.

Answer 2

We know that $e^{i\pi} = -1$.

Transforming: $a^{i\pi/\ln a} = -1$

Then: $a^{(i\pi/\ln a)+\log_a b} = -b$