Solving for an exponent? $1^x = i$
I can't solve it through logs, because $\log 1 = 0$.
Does this mean $x$ is undefined?
 A: As already mentioned in the comments, your problem is not well-posed. Since $i^4 = 1$, you would like to say that $1^{1/4} = i$, but this is not entirely true. In the complex domain, the root function, such as $\sqrt[4]{\cdot}$ is not a well-defined function, because it in general has 4 different values for a single input. It is what is called a multivalued function, or some variant of this.
A: It's true that logarithms are multivalued functions. What this means is that you won't get a unique answer. However certain numbers will clearly work while others do not. What you can do to find solutions is this:
$$1^x=i$$
$$x\ln(1)=\ln(i)$$
We need to find complex values of both of these logarithms. In the following, $n \in \mathbb{Z}$.
$$\ln(1)=a+bi$$
$$1=e^{a+bi}$$
$$1=e^a(\cos(b)+i\sin(b))$$
The imaginary component must be 0 and the real component must be 1. This implies that
$$b=\pi n$$
for any n. This becomes
$$1=e^a(\cos(\pi n)+i\sin(\pi n))$$
$$1=e^a(-1)^n$$
From here it's clear that $a=0$ and $n$ is even, so the full solution for $\ln(1)$ is
$$ln(1)=2\pi n i$$ 
Similarly, with $\ln(i)$:
$$i=e^a(\cos(b)+i\sin(b))$$
Now the real part must be zero
$$b=\frac{\pi (2n+1)}{2}$$
$$i=ie^a(-1)^n$$
So again, $a=0$ and $n$ is even.
$$\ln(i)=\frac{\pi(4n+1)}{2}i$$
Subbing into your equation:
$$x\ln(1)=ln(i)$$
$$x\cdot 2\pi n i=\frac{\pi(4m+1)}{2}i$$
with $m \in \mathbb{Z}$ as well. Solving gives:
$$x=\frac{4m+1}{4n}$$
for any $m,n \in \mathbb{Z}, n \neq 0$.
Saying $\ln(x)$ is undefined in the complex plane is like saying $\sqrt x$ is undefined on the reals. 'Undefined' doesn't mean that no solutions exist, but that multiple solutions exist, meaning that it isn't a function anymore ('multivalued function' is a misnomer since functions by definition only have one output value for a given input). We make it a function by defining a 'principal branch'. For $\sqrt x$ we do this by only considering the positive branch. For complex $\ln(x)$ we do this by restricting the imaginary part of the answer to be in$(-\pi,\pi]$. This would involve dividing by zero here, meaning that the only solutions to your question involve considering non-principal values of the logarithm.
