Cauchy functional equation $f(x+y)=f(x)f(y)$ with $f(2)=5$

Question

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?

Answer 1

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 $$f(5)=25\sqrt{5}.$$

Answer 2

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$.