exp(x) can't reach 0 but exp(exp(x)) can reach 1

If you work with complex numbers and don't just take the principle branch of ln, you can find a solution:

exp(exp(x)) = 1

We can rewrite 1 as exp(2πk*i) with k being any integer. With that we get:

exp(exp(x)) = exp(2πk*i)

Now we can take the complex ln on both sides and get that

exp(x) = 2πk*i

We can multiply any term by 1 without changing the value, and because we know that we can write 1 as exp(2πn*i) with n being any integer, we get:

exp(x) = 2πk*i*1 = 2πk*i * exp(2πn*i)

Now we can take the ln on both sides again.

x = ln(2πk*i * exp(2πn*i))

With the logarithm rule that ln(a*b)=ln(a) + ln(b), we get:

x = ln(2πk*i) * 2πn*i

Because we can't calculate ln(0) we get that k can be any integer but zero.

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Northwest · M

exp(𝑥): it's always positive, and will never reache 0. exp(𝑥) > 0 for all real numbers 𝑥. As 𝑥 approaches negative infinity, exp(𝑥) approaches 0, but never 0. As 𝑥 increases exp(𝑥) grows rapidly.

exp( exp(𝑥)): it is a composition of the exponential function with itself. This means you're taking the exponential of the exponential of 𝑥. Since exp(𝑥) > 0 for any real number 𝑥, exp(exp(𝑥)) is also always positive. Making exp⁡( exp⁡(𝑥)) > 1 for all real numbers 𝑥 because the smallest value exp⁡(𝑥) can take is greater than 0, making exp⁡(exp⁡(𝑥)) > exp⁡(0) = 1.

Here's why "exp(exp(x)) can reach 1" is incorrect:

exp( exp(𝑥)) Cannot Reach 1: The minimum value exp⁡( exp⁡(𝑥)) exp( exp(𝑥)) can take is 1, but it cannot actually be equal to 1. For exp⁡( exp⁡(𝑥)) to be e𝑥actly 1, exp⁡(𝑥) would need to be 0 (since exp⁡(0)=1 exp(0)=1), but exp⁡(𝑥) never reaches 0.

exp(x) Can’t Reach 0: It can get close to 0 as 𝑥 goes to negative infinity, but it never equals 0.

Luke73 · 22-25, M

@Northwest That's true for all real x but as you've seen there's a solution in the complex numbers.

Northwest · M

@Luke73

That's true for all real x but as you've seen there's a solution in the complex numbers.

Sorry, I was not looking through the comments, I was browsing through the headers in the science section. Back to politics. 🤣

Luke73 · 22-25, M

@Northwest I mean everything you said is right lol.

Luke73 · 22-25, M

@helenS Yes but that only works in the complex numbers. And even there exp(x)=0 doesn't have a solution but taking the exp() on both sides and then you have a solution.

helenS · 36-40, F

Isn't that simply because the magnitude of any vector exp(i*π/n) will always be equal to 1? So it can never be the zero vector.

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