Solving f(x^y)=yf(x) | A Functional Equation

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[fix]

This solution is not valid. After defining z as x^y, you can't just say let z=e and we get our solution. x and z are not independent. In fact any group homomorphism (R^+, *)->(R, +) is a solution to this functional equation, the logarithmic function being the most famous one.

[WahranRai]

f(x) = log(x) is a solution of the functional equation because : log(x^y) = ylog(x)
log(x^y) = ylog(x) ----> multiply each side by C (constante): c*log(x^y) = y*c*log(x)
Let g(x) = c*log(x) g(x^y) = y*g(x) ----> solution in the form f(x) = C*log(x) with log of any base

[BoringExtrovert]

For the other comments, the answer k*lnx already contains any base, since the change of base formula is basically only multiplying by a constant

[anotherelvis]

Define g by g(u)=f(e^u) and note f(x)=g(ln(x)). This gives
g(y*ln(x))=y*g(ln(x))
g(y*ln(e))=y*g(ln(e)) Insert x=e
g(y)=k*y
f(x)=k*ln(x) Insert in f(x)=g(ln(x))

[vladislavlukmanov]

I solved that way. x^y=e^(y*ln(x)). On the other hand f(x^y)=y*f(x). So y*ln(x)*f(e)=y*f(x). y cancels out, and we put f(e)=k and then we have: f(x)=k*ln(x).

[SuperPassek]

We can solve it directly: x = e^(ln x), so f(x) = (ln x) f(e) = c ln x where c is a constant.

[zunaidparker]

Here's a 4-line 2AM proof 15 years after I last took calculus, typed out on my phone while in bed. Feel free to point out any mistakes.

Proof:
1. Let f(x^y)=yf(x) be given
2. Observe f(x)=ln(x) is a solution (proven by observation)
3. Let g(x) also be a solution. Then Observe that this ratio is constant for any y, hence g must be a constant multiple of f on all of R
4. Therefore any solution must be a multiple of f, hence all solutions must be of the form f(x)=k.ln(x)

[maciejkowalski]

looks like logarithmic function from the first look

[klausg]

A simple way: let x=e and z=exp(y). Then lhs= f(z) and rhs= lnz • f(e). So f(z)=k•lnz. And it is for all k.

[昆仑云路]

It seems that all log functions fit for it.

[MadScientyst]

As a Real Analysis Major that 'x^y=z, then z=e' kinda looks like a suspicious shortcut.
Although it's a common /useful substitute to cancel each other eg. ln(x).e=1.
Think my Math Professor would've rapped me over if I did that without additional workings.
The proof needs more rigor....IMO

[MrLykhovyd]

@SyberMath actually you have to plug it back and check that it is an actual solution. First steps assumed that such function f exists and then you did substitutions like taking the value of f(e). I vaguely remember there might be exotic cases when you can derive f(x) and substitution back to the original equality fails.

[IlaiShoshani]

I came up with an equation but I don't know if there is a way to solve it:

f(x)f(-x) = f'(x)f'(-x)

The solution is nice tho

[zunaidparker]

This "proof" seems completely circular and a bit too hand wavey. You use log functions to simplify your equation then brute force your choice of answer to give you a log function as the outcome?
Your answer is correct in the end but this is not a proof by any means that would get you marks in an exam.
I think blackpenredpen or Michael Penn would do such a problem statement justice with a proper proof.

[YahiaNebti]

What about log, I think its with any base

[sil]

Or... just let directly x=e, y=ln z for arbitrary z>0 and you get f(z)=f(e)*ln z=k*ln z. Also I am not sure why you had the continuity assumption, it's not used anywhere, this is the necessary solution regardless.

[Jono]

Actually, logarithms of any base would be a solution to this functional equation, not just the ln function.