How can one solve $1^x=2$?

Question

Sure, common sense says there's no solution. But, I feel, there should be one! (If there isn't, can't we construct one?)

Answer 1

There is no real solution for the first question $1^x = 2$

For the second part:

Let $x = a+ib$ and $e^{(2n+1)i\pi} = -1 , n \in \Bbb Z $ so, $2^{a+ib} = 2e^{(2n+1)i\pi}$, so $2^{a-1 + ib} = e^{(2n+1)i\pi}$.

Take log and you will get, $(a-1+ib)\log2 = (2n+1)i\pi$.

Equating real and imaginary part, $a = 1, b = \frac{(2n+1)\pi}{\log2}$. Thus

$$x = 1 + i\frac{(2n+1)\pi}{\log2}, n \in \Bbb Z.$$