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Let $f$ be any function defined on set $\Bbb N$.if $f(xy)=f(x)+f(y)$ & $f(2)=9$ then find $f(3)$ and answer for this question is $7$ . Please say how to get it.

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If $f$ is continuous, we have $f(x)=\log_a(x)$ for $a$ such that $a^9=2$. So $a>1$ and $f(x)$ is monotonically increasing and we cannot t have $f(3)=7<9=f(2)$.

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  • $\begingroup$ It is only defined for positive integer. No question of continuity. $\endgroup$ – Arindam Mandal May 5 '18 at 13:37
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Your claim is not true. For instance, $f(x)=9(a_{p_1}+\cdots+a_{p_h})$, where $x=p_1^{a_1}\cdots p_h^{a_{p_h}}$ is its prime factorization, satisfies $f(xy)=f(x)+f(y)$ and $f(2)=9$. Yet, $f(3)=9$. Specifically, for all functions $g:\{\text{primes}\}\to \Bbb N$, there is exactly one function $f:\Bbb N\setminus\{0\}\to\Bbb N$ such that $f(p)=g(p)$ for all primes and $f(xy)=f(x)+f(y)$ for all $x,y$.

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  • $\begingroup$ Not clear to me. $\endgroup$ – Arindam Mandal May 5 '18 at 17:12
  • $\begingroup$ What is not clear? $\endgroup$ – user228113 May 5 '18 at 17:13
  • $\begingroup$ No where it is said that f(3)=9.. $\endgroup$ – Arindam Mandal May 5 '18 at 17:15
  • $\begingroup$ I don't (nor can) know why your assignment is wrong, but fact is that a function $f$ such as yours cannot be determined uniquely by the value it takes at $x=2$. $\endgroup$ – user228113 May 5 '18 at 17:18

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