Attempt to formally generalize log, exp functions to 3,4,5..(n,m) log exp

[bo198214]

So, not only can we show that f(x) is a constant, but we can show it in multiple ways, which is a very bad sign. I think this is proof that no such function can exist (or the only solution to all of these equations is f(x)=1).

Yeah, I see its even much simpler to prove than I did, and we can also drop the demand of continuity. However your proof(s) omits the case of \( f(x)=0 \). So I will add the detail:

If \( f(x^y)=f(x)f(y) \) for f being defined on \( [1,\infty) \), then \( f(x)=f(x)f(1) \) for \( y=1 \). So if there is at least one \( x \) with \( f(x)\neq 0 \) (i.e. \( f \) is not identically 0) then \( f(1)=1 \).
Let now \( x=1 \) in our original equation: \( f(1)=f(1)f(y) \). With \( f(1)=1 \) we get \( f(y)=1 \) for all \( y \).

So either \( f=1 \) or \( f=0 \).

@Ivars: this proof also works for hyperreals.