Why $\Re\left(e^{\sqrt{-\log x}}\right)=\Im\left(e^{\sqrt{-\log x}}\right)$ has no real solutions?

Question

I don't know if next question is obvious, any case I am asking about it. I consider the equation over complex numbers $$\Re\left(e^{\sqrt{-\log x}}\right)=\Im\left(e^{\sqrt{-\log x}}\right),$$ where I understand the definiton of this input in Wolfram Alpha online calculator

solve Re(e^(sqrt(-log(x))))=Im(e^(sqrt(-log(x))))

Question. I don't know if there is an obvious/quick way to deduce that the equation $$\Re\left(e^{\sqrt{-\log x}}\right)=\Im\left(e^{\sqrt{-\log x}}\right)$$ has no real solutions. Any case, please provide hints or a solution to deduce that there is no real solutions. Many thanks.

Thus is required that the definition of the square root and logarithm is the used by the the online calculator, I presume that the usual ones.

Answer 1

There are real solutions. One is given by $$ x = e^{\pi^2/16}, $$ since $\log{x} = \pi^2/16$, so $\sqrt{-\log{x}}=\pm i \pi/4$. If, as is conventional, we take $+i$ as the principal square root of $-1$, we then have $$ e^{\sqrt{-\log{x}}} = e^{i\pi/4} = \cos{(\pi/4)}+i\sin{(\pi/4)} = \frac{1+i}{\sqrt{2}}, $$ which has equal real and imaginary parts. If we take the other square root, $x=49\pi^2/16$ works.

In general, in the first case there are roots at $x=e^{\pi^2 (2n+1/4)^2}$, in the second at $x=e^{\pi^2(2n-1/4)^2}$ for integer $n$.

Answer 2

You should never assume that an equation has no solution just because Alpha tells you there isn't one. Instead, it's more likely that its rules for transforming equations don't allow it to bring this one into a form where it can do anything with them. Instead, let's try to solve it directly. Saying that $\Re(z)=\Im(z)$ is exactly the same as saying that there exists $a\in\mathbb{R}$ with $z=ae^{i\pi/4}$ (i.e., that there's some $b\in\mathbb{R}$ with $z=b+bi$). Now, we can work backwards: $e^{\sqrt{-\log x}}=ae^{i\pi/4}$, so $\sqrt{-\log x}=i\pi/4+\log a$ (be careful — I'm being a little cavalier with the branches of logs here!), so $-\log x = (\log^2a-\pi^2/16)+(\log a)i\pi/2$, so $x = e^{\pi^2/16-\log^2a}e^{-(\log a)i\pi/2}$.