# Does a function that has an exponential analog to $log(xy) = log(x) + log(y)$ exist?

Similar to how $log(xy) = log(x) + log(y)$, does a nontrivial function exist that has the property $f(x^y) = f(x)f(y)$? How would one attempt to derive such a function?

• I don't understand the analogy here? – copper.hat Nov 15 '13 at 4:19
• @copper.hat I guess the idea is to replace multiplication with exponentiation and addition with multiplication. – Karl Kroningfeld Nov 15 '13 at 4:21
• Exponentiation is not analogous to multiplication. The latter is symmetrical algebraic operation, the former is not. – anon Nov 15 '13 at 4:22

Then $f(x^y) = f(y^x) \Rightarrow f(1^y) = f(y^1) \Rightarrow f(1) = f(y)$ for any $y$. f is a constant function, $0$ or $1$.
• You missed an $f$. – Bennett Gardiner Nov 15 '13 at 4:48
• Just a little typo: cons${\Large\bf\mbox{t}}$ant. It's a fine proof. UpVote $0$ k. – Felix Marin Nov 15 '13 at 5:10