Find the function(s) satisfying $f(x/y)=f(x)-f(y)$ and $f(e)=1$ for $x>0$.
Although $\ln x$ is the obvious solution to the above problem, I am interested in knowing the general way to deal with such problems(in case the function is not so obvious)
My approach:
$f(x/y)=f(x)-f(y)$
Substitute $y=1/x$
So, $f(x^2)+f(1/x^2)=0$
How do I proceed further?
On
@KaviRamaMurthy already showed that the problem can be reduced to
Find the functions $g:\mathbb R \to \mathbb R$, such that $g(x+y)=g(x)+g(y)$ and $g(1)=1$!
We will first show that $g$ is uniquely defined on $\mathbb Q$. Then we will show that there is only one such $g$ on $\mathbb R$ if $g$ is continuous.
We have
$$g(0)=g(0+0)=g(0)+g(0)$$ and therefore $g(0)=0$. Further we have $$0=g(0)=g(n+(-x))=g(x)+g(-x)$$ From this follows $g(-x)=g(x)$. By induction we prove
$$g(nx)=ng(x), \forall n \in \mathbb N, \forall x \in \mathbb R$$ and with the previous results this gives $$g(nx)=ng(x), \forall n \in \mathbb Z, \forall x \in \mathbb R$$ If we set $x=1$ this means $$g(n)=n, \forall n \in \mathbb Z, $$ From $$1=g(1)=g(n\cdot \frac 1 n)=ng(\frac 1 n)\forall n \in \mathbb N$$ follows $g(\frac 1 n)=\frac 1 n,\forall n\in \mathbb N$ and so $g(\frac m n)=\frac m n,\forall m \in \mathbb Z, \forall n \in \mathbb N$. But this means $$g(x)=x, \forall x \in \mathbb Q$$ This proves our first statement.
If $x\in \mathbb R$, then there exist a sequence $x_k \in \mathbb Q,k \in \mathbb N$, such that $$\lim_{k\to\infty}x_k=x$$ If $g$ is continuous, then $$g(x)=g(\lim_{k\to\infty}x_k)=\lim_{k\to\infty}g(x_k)=\lim_{k\to\infty}x_k=x$$ and so $$g(x)=x,\forall x \in \mathbb R$$ Note note that $g(qx)=qg(x)$ for a $q\in \mathbb Q$. Also that if $g$ is continuous in at least one point the $g$ is continuous on $\mathbb R$.
So if there are other solutions to this functional equation then they are not continuous at any point.
How does such other solution looks like?
If $q \in \mathbb Q$ then $g(qx)=qg(x)$, even if $x \in \mathbb R\setminus \mathbb Q$.
A set $A\subset \mathbb R$ is independent : $$\forall \{a_1,\ldots,a_k\} \subseteq A,\;\forall \{q_1,\ldots,q_k\} \subseteq \mathbb Q:\\ q_1 a_1+\ldots+q_k a_k=0 \implies q_1=\ldots=q_k=0$$
A set $A\subset \mathbb R$ is a generating system:
$$\forall r \in \mathbb R, \; \exists \;\{a_1,\ldots,a_k \} \subseteq A,\; \exists\{q_1,\ldots,q_k\}\subseteq \mathbb Q:\\ r=q_1 a_1+\ldots+q_k a_k$$
$A$ is independent and generating system, and $h:A->\mathbb R$
If $r \in \mathbb R$ then $r$ can be uniquely represented as $r=q_1 a_1+ \ldots+ q_k a_k$, and we define
$$g(r)=g(q_1 a_1+ \ldots+ q_k a_k):=q_1 h(a_1)+\ldots+q_k h(a_k)$$
This function $g$ has the additivity property
$$g(a+b)=g(a)+g(b), \forall a,b \in \mathbb R$$
The proof is simple. Assume that
$$a=q_1 a_1 + \ldots +q_i a_i + s_1 c_1 + \ldots +s_k c_k $$ and $$b=p_1 b_1 + \ldots +p_j b_j + t_1 c_1 + \ldots + t_k c_k$$
such that
$$\{a_1,\ldots,a_i,c_1,\ldots, c_k, b_1,\ldots b_j\}\subseteq A$$ and
$$ \{q_1,\ldots,q_i,s_1,\ldots, s_k, p_1,\ldots p_j\}\subseteq \mathbb Q$$
Then
$$g(a + b)\\= g(q_1 a_1 + \ldots +q_i a_i + s_1 c_1 + \ldots +s_k c_k +
\\ p_1 b_1 + \ldots +p_j b_j + t_1 c_1 + \ldots + t_k c_k)
\\= g(q_1 a_1 + \ldots +q_i a_i +\\
(s_1+t_1) c_1 + \ldots + (s_k + t_k) c_k \\
+ p_1 b_1 + \ldots p_j b_j )
\\= q_1 h(a_1) + \ldots + q_i h(a_i) \\
+(s_1+t_1) h(c_1) + \ldots + (s_k + t_k) h(c_k) \\
+ p_1 h(b_1) + \ldots p_j h(b_j)\\
=q_1 h(a_1) + \ldots + q_i h(a_i) + s_1 h(c_1) + \ldots + s_k h(c_k) \\
+ p_1 h(b_1) + \ldots p_j h(b_j) + t_1 h(c_1)
+ \ldots + t_k h(c_k)
\\=g(q_1 a_1 + \ldots q_i a_i + s_1 c_1 + \ldots + s_k c_k) \\
+g(t_1 a_1 + \ldots q_i a_i + s_1 c_1 + \ldots + s_k c_k)
\\= g(a)+g(b)$$
Of course we can adjust $f$ such that $g(1)=1$.
I do not want to prove here that such a set $A$ exists in $\mathbb R$.
This proof and a proof that such a set $A$ exists is published in Eine Basis aller Zahlen und die unstetigen Loesungen der Funktionalgleichung: f(x+y)=f(x)+f(y) by Georg Hamel, Mathematische Annalen, Leipzig,1906
There is no such thing as a general way of solving functional equations.
Let $g(x)=f(e^{x}), x \in \mathbb R$. Changing $x$ to $xy$ in the given equation we get $f(xy)=f(x)+f(y)$ and this becomes $g(x+y)=g(x)+g(y)$. There are lots of weird solutions of this equation but the only continuous one with $g(1)=1$ is $g(x)=x$. This gives $f(x) =\ln x$ if continuity is assumed.