What is the general solution to equations of form $f(k\cdot x)-lf(x)=1, k,l \in \Bbb N$ ? Prove that $\log_{10} x$ is the solution when $k = 10,l =1$. I have a function $f:\mathbb{R}^{>0}\rightarrow \mathbb{R}$ that satisfies the following equation:
$$f(10\cdot x)-f(x)=1 \quad\quad (1)$$
From looking at this equation I can easily see that $f(x)=\log_{10}(x)$ works out nicely.
However, if the equation would have been 
$$f(10\cdot x)-5f(x)=1 \quad\quad (2)$$
then the logarithm would not have worked out. 
Two questions:


*

*How can I prove that the logarithm is the only solution to equation (1)?

*Is there a general way to solve these kind of equations, for example, how would one solve equation (2)?

 A: For the first question, $\log_{10}$ is not the only solution: let $g$ be any function defined on $[1, 10)$. Define
$$f(x) = k + g(x \cdot 10 ^{-k})$$
when $10^k \le x \lt 10^{k+1}$. Then $f$ is a solution of the functional equation. Note that $k = \lfloor\log_{10}(x)\rfloor$
For the second equation, you can use a similar definition, I suggest
$$f(x) = 5^k \left( g(x \cdot 10 ^{-k}) + \frac{1}{4}\right) - \frac{1}{4}$$
A: As Gribouillis points out, there are no unique solutions to equations like this.
The equation only connects the values of $f(x)$ for values of $x$ that differ by factor ten, so it says nothing about $f(x)$ and $f(y)$ when $x/y$ is not an integer power of ten.
However, there is a way to find a "nice" solution to the kind of equation.
If you want the general solution instead of pretty ones, this is not your answer.
By pretty I mean something that satisfies a natural generalization of your condition.
Consider the second problem.
Take any $x\in(0,\infty)$ and $n\in\mathbb N$.
We have
$$
\begin{split}
f(10^nx)
&=
1+5f(10^{n-1}x)
\\&=
1+5(1+5f(10^{n-2}x))
\\&=
(1+5)+5^2f(10^{n-2}x)
\\&=
\dots
\\&=
(1+5+25+\dots+5^{n-1})+5^nf(x)
\\&=
\frac14(5^n-1)+5^nf(x)
\\&=
5^n(\frac14+f(x))-\frac14
.
\end{split}
$$
This formula can be used to find the general solution if you want to.
To find a pretty solution, we add the additional restricting assumption that this holds for all values of $n\in\mathbb R$.
For $x=1$ and $f(1)=a-1/4$ (the shift is for convenience) this leads to
$$
f(y)
=
f(10^{\log_{10}y})
=
%5^{\log_{10}y}(\frac14+a)-\frac14
5^{\log_{10}y}a-\frac14
.
$$
It is easy to check that this indeed satisfies
$f(10^tx)
%=5^{t+\log_{10}x}a-\frac14=5^t5^{\log_{10}x}a-\frac14
=5^t(f(x)+1/4)-1/4$
for all $x>0$ and $t\in\mathbb R$.
You are free to choose the parameter $a$.
For $a=0$ you get the constant function $f(x)=-1/4$.
Similarly, for the first problem you would get $f(y)=a+\log_{10}y$ with this method.
A: $$f(10\cdot x)-f(x)=1 \quad\quad (1)$$
By induction it's easy to prove $ f(10^n) = n + f(1) $
So any function $g(s) \longrightarrow g(1) +  \log_{10} $(s) satisfy $(1)$
With $ s \ne 0 $ ....
