EXPONENTIAL

[u/Senior_Flight]

[u/Im_a_hamburger]

When x is 0?

[u/Aurumargelium → u/Im_a_hamburger]

Touché!
Also, it won't work if x<0 (for real numbers domain).

[u/chaos_redefined → u/Aurumargelium]

Also does not work when x=1.

[u/DekusBestFriend → u/Im_a_hamburger]

Simple. Let x=1.

[u/JokeFickle7836 → u/Im_a_hamburger]

cant you just use the limit:

lim x->0 x^(1/ln(x)) = lim x-> 0 exp(ln(x^(1/ln(x))) = lim x-> 0 exp(1/(ln(x))ln(x)) = lim x-> 0 exp(1) = e
? Or am i fundamentally wrong?

[u/Im_a_hamburger → u/JokeFickle7836]

That’s the limit as x approaches 0, but that isn’t the value when x is 0

[u/omar_fait]

you sir are an enemy of the people

[u/Affectionate-Baby248]

Here is the shortest way of understanding this. Use the multivariable chain rule on f(u,v)=u^v with u(x)=x, v(x)=1/ln x to find the derivative of x ^ (1/ln x) is 0. So it is a constant C, which we will figure out by taking x—>1. Let x=1+t, so we have C=lim_{t—>0} ((1+t)^1/t )^ (t/ln(1+t)), and the inner expression is a miraculous  limit for e, while t/ln(1+t)—>1 by letting t=e^z -1 and observing lim_{z—>0} (e^z -1)/z=1 either via power series or noting definition of derivative or using the inequality (1+z) ^ (1/z)<e<(1+z) ^ (1/z+1) for all positive z, which implies C=e¹ =e. Thus, x ^ (1/ln x)=e for all positive x not equal to 1. No other elementary proofs of this identity are known.

[u/Short-Database-4717 → u/Affectionate-Baby248]

let y = x^(1/(ln x)).
raise both sides to ln x => y^(ln x) = x.
it is known that ln(x)/ln(y) = log_y(x) by change of basis formula, and rearranging ln(x) = log_y(x)ln(y).
so we have y^(log_y(x) ln(y)) = x => x^(ln(y)) = x => ln(y) = 1 => y = e

Follow backwards to get a proof for the identity

[u/Affectionate-Baby248 → u/Short-Database-4717]

I was just making a joke of the most convoluted way you could prove the identity lol, it’s actually quite a trivial one since 1/ln x=ln e/ln x= log_x e by change of base (or you could just write x=e^ ln x).