A monotonic multiplicative integer functional equation. Let $ f:\mathbb N \to \mathbb N $ be such that


*

*$ f (x)> f (y)$ if $x> y$.

*$ f (xy)=f (x) f (y) $.

*$ f (3) \geq 7$.
Find the smallest value of $ f (3) $
My attempt:if we can define the function for all primes. But if we define them unsuccessfully we would end up functions that contradicts condition one:the monotonicity condition. 
 A: We have that $f(2^5)=f(2)^5 > f(3^3)\geq 343\Rightarrow f(2) \geq 4 $. Since $3^5 = 243$
On the other hand we have 
$f(9)= f(3)^2 > f(2)^3\geq 64\Rightarrow f(3) >8$.
Combined with Myerson's comment we conclude $f(3)=9$. Since $f(n)=n^2$ says the least possible  $f(3)$ is at most $9$.
A: Here's another strategy.  It is less clever, but offers a little more generality.  Let $p$ be another natural number and let the map $m$ (associated to $p$) be defined on the natural numbers by $$ m(k) = \max_{l \geq 0} \{ 3^k \geq p^l \}. $$  Then we have 
$$p^{m(k)} \leq 3^k \leq p^{m(k) + 1}. $$
In particular, for all $k$ in $\mathbb{N}$, 
$${m(k) \over k} \leq {\log(3)\over \log(p)} \leq {m(k)+1 \over k}$$
so
$$ 0 \leq {\log(3) \over \log(p)} - {m(k) \over k} \leq {1\over k} \to 0$$
i.e. ${m(k) \over k} \to {\log(3) \over \log(p)}$.  By the conditions on $f$ and analogous steps we find
$$0 \leq {\log(f(3)) \over \log(f(p))} - {m(k) \over k} \leq {1\over k} \to 0$$
and so conclude that ${m(k) \over k} \rightarrow {\log(f(3)) \over \log(f(p))}$ as $k$ goes to $\infty$.  It follows by uniqueness of limits that ${\log(3) \over \log(p)} = {\log(f(3)) \over \log(f(p))}$ giving
$$7^{\log(p)} \leq f(3)^{\log(p)} = f(p)^{\log(3)}.$$ This gives a lower bound on $f$ everywhere.  Note that nothing new is achieved if $p=3$, so we appeal to a lower integer.  In particular $f(2) \geq 7^{\log(2)\over \log(3)} \approx 3.4$ and so $f(2) \geq 4$.  Using the above display again,
$$ f(3)^{\log(p)} = f(p)^{\log(3)} \geq 4^{\log(3)} $$
leading to the estimate $f(3) \geq 4^{\log(3) \over \log(2)} = 9$.
It is interesting to note that if we choose $f(3) = 3^n$, then $f(p) = p^n$ ($n \geq 2$).  It appears that these are the only possible functions.
