What is the set of numbers that are roots of the exponential?

Question

Usually when an equations has no roots it leads to a new set of numbers. For example $x^2+1=0$ lead to the development of complex and imaginary numbers. What are the extension of numbers that solve $$e^x=0\;?$$ Obviously this is not possible for $x\in\mathbb{C}$. I am having a bad time looking for this as many websites are dedicated to finding the roots of equations with exponentials. I am guessing since

$$\log(e^x)=x = \log(0) (?)=-\infty$$

it is somehow related to hyperreal numbers?

In the affinely extended real number line, log(0) is defined as -∞ (just in case you don't know, the affinely extended real number line includes ∞ and -∞ as real numbers). — MathGeek, Aug 23 '21 at 11:30
See the sci.math thread Is e^x zero in some extended number system? I started in 30 January 2006 (5 posts, all made the same day). Interestingly, the paper I discuss seems to be freely available on the internet (I thought it would be behind a paywall): Paul Dienes, The exponential function in linear algebras, Quarterly Journal of Mathematics (Oxford) 1 (1930), pp. 300-309. — Dave L. Renfro, Aug 23 '21 at 12:20
Somewhat related: https://math.stackexchange.com/questions/259584/why-dont-we-define-imaginary-numbers-for-every-impossibility — Hans Lundmark, Aug 23 '21 at 13:06

score 1 · Accepted Answer · answered Aug 23 '21 at 11:40

If you are working with the hyperreal numbers, log(0) ≠ −∞, but if you are working with the affinely extended real numbers, log(0) = −∞, so to answer your question, the extension of numbers that solve e^x = 0 are the affinely extended real numbers.

score 1 · Answer 2 · answered Mar 28 '24 at 16:34

1

In extended real numbers $\overline {\mathbb R}$, the solution for $e^x=0$ is $-\infty$. Extending reals with logarithm of zero can be done in a different way as well.

answered Mar 28 '24 at 16:34

Anixx

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What is the set of numbers that are roots of the exponential?

2 Answers2