## Task 20/80

We are looking for a set of functions $f$ that satisfy the equation $f(x+1)=c \cdot f(x)$ for every real number $x$ (where $c$ is a positive real constant).
If $f(x+1)=c \cdot f(x)$ is to hold for every real number $x$, then the equation must specifically hold for $x=0 ; 1 ; 2 ; 3 ; \ldots$. Therefore,

\[
\begin{array}{r}
f(1)=c \cdot f(0) \\
f(2)=c \cdot f(1)=c^{2} \cdot f(0) \\
f(3)=c \cdot f(2)=c^{3} \cdot f(0) \\
\cdots \\
f(n)=c \cdot f(n-1)=c^{n} \cdot f(0)
\end{array}
\]

It is conjectured that $f(x)=c^{n} \cdot f(0)$. Indeed, it holds that

\[
f(x+1)=c^{x+1} \cdot f(0)=c \cdot c^{x} \cdot f(0)=c \cdot f(x)
\]

With $f(0)=a$, the desired set $M$ is

\[
M=\left\{f \mid f(x)=a \cdot c^{x} \text { with } a ; c ; x \in P \text { and } c>0\right\}
\]

(where $P$ denotes the set of real numbers).

Postscript: It has not been shown that no other set exists besides the one specified; however, this is not required by the problem statement!