Solve for continuous function $f$ where $f(bx)-f(by)=f(x)-f(y)$ for all positive number $b$. $f$ is defined on positive numbers.
Example: $f(x)=\log x$
Check: $f(bx)-f(by)=\log x+\log b-\log b-\log y$ (verified).
My try: first consider the case $f(1)=0$. Since $f(bx)-f(by)=f(x)-f(y)$, then $f(xy)-f(y)=f(x)-f(1)$
Then $f(xy)=f(x)+f(y)$
Let $g(x)=f(e^x)$
Then $g(x+y)=g(x)+g(y)$ this is the Cauchy equation. Therefore $g$ is linear and $f$ is logarithm.
Am I on the right track?
Answer: $f(x) = A \ln x + B$.
Proof.
Case $f(1) = 0$ is already done by High GPA in post. If $f(1) \ne 0$, consider $F(x) = f(x) - f(1)$. Hence $F(bx) - F(by) = F(x) - F(y)$ and we get the case with $F(1)=0$. As it was already proved, $F(x) = const * \ln x$. Hence $f(x) = F(x) + f(1) = A \ln x + B$. We showed that $f(x) = A \ln x + B$. It's a necessary condition. It's obvious, that it's also sufficient. The problem is solved.
P.S. I deleted the previous long proof.