Cauchy functional equation $f(x+y)=f(x)f(y)$ with $f(2)=5$ 
If $f$ is a real-valued function with $x,y \in \mathbb{R}$ such that
  $$f(x+y)=f(x)f(y)$$
  then find $f(5)$, given that $f(2)=5$.

So, can someone tell me if/where I'm incorrect? This was my approach:
$$f(1)f(1)=f(1+1)=f(2)=5$$
$$f(1)=\sqrt5$$
$$f(2)f(2)=5^2=f(2+2)=f(4)$$
$$f(4)=25$$
In general, we have 
$$f(2x)=f^2(x)$$
by substituting $y=x$. Combining these results, we reach
$$f(5)=f(4+1)=f(4)f(1)=25\sqrt5$$
I saw this problem on a website, and my answer was marked incorrect. Someone care to clarify?
 A: This functional equation has name the exponential
Cauchy functional equation and a real-valued function is called a real
exponential function if it satisfies this functional equation. The general solution of the exponential Cauchy functional equation is given by 
$$f(x)=e^{A(x)}\ \ \ \ and\ \ f(x)=0$$
where $A:\mathbb{R}\rightarrow \mathbb{R}$ is an additive function and e is the Napierian base of
logarithm. For proof of this result and much more related results please see chapter $1$ and $2$ of the following book.
${Prasanna\ K.\ Sahoo\ and\ Palaniappan\ Kannappan,\ "Introduction\ to\ Functional\ Equations",\ 2011\ by\ Taylor\ and\ Francis\ Group,\ LLC.}$
Hint. $f(2)=5$, so with the above result there is an additive function $A:\mathbb{R}\rightarrow \mathbb{R}$;
$$f(x)=e^{A(x)};$$
$A(2)=\ln 5$, since $A$ is additive $A(1)=\ln \sqrt{5}$ an so $A(5)=5\ln \sqrt{5}$, thus 
$$f(5)=25\sqrt{5}.$$
A: The error is in assuming that $f(1)^2=5 \implies f(1)=\sqrt 5$ ( and not $(-\sqrt 5$).You can correct it by beginning with $ [ \forall x (  f(x)=f(x/2+x/2)=f(x/2)^2 \ge 0 ) ] \implies \forall x ( f(x) \ge 0 )$. Footnotes:(1) It is better to write $f(x)^2$ than $f^2(x)$ as this can be read as $f(f(x))$ even though we usually make an exception to this with trig functions.(2) The function $A(x)$ in Deliasaghi's answer cannot be assumed to be $kx$ for constant $k$ unless $A(x)$ is assumed to be continuous. There are (many) discontinuous additive $ A : R \to R$.  
