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.

Luke73 · 26-30, 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.