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How many functions satisfy this set of conditions?

  1. Sep 14, 2011 #1
    How many functions satisfy this set of conditions? 1) f(x.y) = f(x) + f(y). 2) f(a)=1.
    I was trying to define a logarithmic function abstractly as a map from R+ to R satisfying those 2 conditions that this question came to my mind. It's obvious that there are infinitely many functions satisfying the 1st condition (each logarithmic function in an arbitrary base satisfies the first condition). but I hope that the 2nd condition could be used to prove that only one fixed function satisfies both conditions (namely, I want to prove that only loga satisfies both conditions and then I'll use this idea to abstractly define the map loga: R+ -> R).
    Is there any way to prove or disprove this conjecture? I've been trying to prove it from tomorrow morning and have failed so far. my idea is to define h(x) = f(x) - g(x) where f and g are functions with the same domain satisfying those 2 conditions and then to show that h(x)=0 for all x's in the domain.
  2. jcsd
  3. Sep 14, 2011 #2
    What exactly do you require of your function f? I'm assuming it is real-valued, but what set is it from? You example of log is just a function [itex]\mathbb{R}^+ \to \mathbb{R}[/itex]. Are we just looking for a function [itex]A \to \mathbb{R}[/itex] where A is any set of real numbers that contains a and is closed under multiplication? Furthermore do you require f to be continuous?

    One approach you may try is to define [itex]F : \mathbb{R} \to \mathbb{R}[/itex] by
    [tex]F(x) = f(e^x)[/tex]
    F will then satisfy Cauchy's functional equation (look it up on e.g. Wikipedia if you have never heard of it) which should help you conclude some stuff about the solutions. In particular if you require some nice conditions like continuity at a point you will get a unique solution, but otherwise you will have an infinite family of solutions.
  4. Sep 14, 2011 #3


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    If you are using the axiom of choice there are infinitely many and all of them except log are really horrible. You probably want some other condition on f. Many would work but the standard choice is
    for all 0<x,y
    1) f(x.y) = f(x) + f(y). 2) f'(1)=1.
    There is only one such function f(x)=log(x)
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