Real Analysis. Prove f(x) = logx given all these conditions.

[harmonie_Best]

It's just the final part (e) that I don't get, I have the rest but just for completeness I thought I'd put it in

(iii) Let f : (0,infinity) -> R be a function which is differentiable at 1 with f '(1) = 1
and satisfies:

f(xy) = f(x) + f(y) (*)

(a) Use (*) to determine f(1) and show that f(1/x) = -f(x) for all x > 0.
Got that f(1) = 0 and proved the second part

(b) Use (*) to show that f is differentiable at a with f '(a) = a^(-1) for all a > 0.
Yep

(c) State the mean value theorem.
Yep

(d) Use the mean value theorem to prove that a differentiable function g :(0,infinity) -> R with g'(x) = 0 for all x in (0,infinity) must be constant.
Basically constant value theorem.

(e) Apply the previous part to show that f(x) = log x for all x.
You can use the fact that log x is differentiable on (0,infinity) and has log' x = x^(-1) for all x in (0,infinity)

Sorry if this seems long and too easy but I just don't get how you would implement (d) (constant value theorem) to get what you
want? I would have thought you would use (b)

Cheers

 

[sutupidmath]

I would say define a helping function g as follows:

$$g(x)=f(x)-log(x)$$, then $$g'(x)=0$$ so g is constant. So $$f(x)-log(x)=K$$. Now you can determine K from previous parts.

Does this help?

This is a nice problem by the way, I remember doing it through a different approach some time ago.

 

[harmonie_Best]

Perfecto! Thanks a bunch!