MHB Determine all real valued differentiable functions f(x)+f(y)=f(xy)

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The discussion focuses on finding all real-valued differentiable functions f defined for x > 0 that satisfy the equation f(x) + f(y) = f(xy) for all x, y > 0. By setting x = y = 1, it is established that f(1) = 0. Differentiating the equation with respect to x leads to the conclusion that xf'(x) is constant, allowing for the integration to yield f(x) = c ln x + d. Given that f(1) = 0, it follows that d must equal 0, resulting in the final solution f(x) = c ln x. These functions are the only solutions to the problem.
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Determine, with proof, all the real-valued differentiable functions $f$, defined
for real $x > 0$, which satisfy $f(x) + f(y) = f(xy)$ for all $x, y > 0$.
 
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The first thing I would notice is that, taking y= 0, f(x)+ f(0)= f(0) so that f(x)= 0 for all x.
 
because f(xy) = f(x) + f(y) and defined for x, y >0 I see a log function $f(x) =m\log\,x$ satisfies criteria where m is arbitrary constant. m= 0 gives above solution(post #2)
 
lfdahl said:
Determine, with proof, all the real-valued differentiable functions $f$, defined
for real $x > 0$, which satisfy $f(x) + f(y) = f(xy)$ for all $x, y > 0$.
[sp]With $x=y=1$, $f(1) + f(1) = f(1)$, from which $f(1) = 0$.

Fix $y$, and differentiate the equation $f(x) + f(y) = f(xy)$ with respect to $x$: $f'(x) = yf'(xy)$. Therefore $xf'(x) = xyf'(xy)$.

Now let $z = xy$, so that $xf'(x) = zf'(z)$. Since that is true for all positive numbers $x$ and $z$, it follows that $xf'(x)$ is constant, say $xf'(x) = c$. Then $f'(x) = \dfrac cx$, so we can integrate to get $f(x) = c\ln x + d$ (where $d$ is another constant). But $f(1) = 0$, so that $d=0$. Thus $f(x) = c\ln x$, and those are the only solutions.[/sp]
 
kaliprasad said:
because f(xy) = f(x) + f(y) and defined for x, y >0 I see a log function $f(x) =m\log\,x$ satisfies criteria where m is arbitrary constant. m= 0 gives above solution(post #2)
Hi, kaliprasad!
Your intuitive solution is correct indeed! Thanks for participating!

- - - Updated - - -

Opalg said:
[sp]With $x=y=1$, $f(1) + f(1) = f(1)$, from which $f(1) = 0$.

Fix $y$, and differentiate the equation $f(x) + f(y) = f(xy)$ with respect to $x$: $f'(x) = yf'(xy)$. Therefore $xf'(x) = xyf'(xy)$.

Now let $z = xy$, so that $xf'(x) = zf'(z)$. Since that is true for all positive numbers $x$ and $z$, it follows that $xf'(x)$ is constant, say $xf'(x) = c$. Then $f'(x) = \dfrac cx$, so we can integrate to get $f(x) = c\ln x + d$ (where $d$ is another constant). But $f(1) = 0$, so that $d=0$. Thus $f(x) = c\ln x$, and those are the only solutions.[/sp]

Thanks, Opalg! for your participation. Your solution is - of course - correct!(Yes)
 
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