How do I prove that $f(x)f(y)=f(x+y)$ implies that $f(x)=e^{cx}$, assuming f is continuous and not zero? This is part of a homework assignment for a real analysis course taught out of "Baby Rudin."  Just looking for a push in the right direction, not a full-blown solution.  We are to suppose that $f(x)f(y)=f(x+y)$ for all real x and y, and that f is continuous and not zero.  The first part of this question let me assume differentiability as well, and I was able to compose it with the natural log and take the derivative to prove that $f(x)=e^{cx}$ where c is a real constant.  I'm having a little more trouble only assuming continuity; I'm currently trying to prove that f is differentiable at zero, and hence all real numbers.  Is this an approach worth taking?
 A: Since $f(t)=f(t+0)=f(t)f(0)$ we can conclude that $f(0)=1$.  If there exists a real number $β$ such that $f(β)=0$, then for any real number $x$ we have 
$f(x)=f(β+(x-β))=f(β)f(x-β)=0$
which implies that $f$ is identically zero.  By the continuity of $f$ we can conclude that $f(x)>0$ for all real numbers.
For a positive integer $n$, $f(n)=f(1+1+...+1)=f(1)f(1)*...*f(1)=f(1)^n$.
Since $f(-1)=f(1+(-2))=f(1)f(-2)=f(1)f(-1)f(-1)$,
it follows that $f(-1)=f(1)^{-1}$.
For a negative integer $m$, $f(m)=f(-1+(-1)+...+(-1))=f(-1)^{-m}=(f(1)^{-1})^{-m}=f(1)^m$.
Since $f(0)=1=f(1)^0$, $f(n)=f(1)^n$ holds for all integers.
Let $q$ and $m$ be positive integers, then $f(m)=f(1/q+1/q+...+1/q)=f(1/q)^{qm}=f(1)^m$.
It follows that $f(1/q)=f(1)^{1/q}$.  This will also hold for negative integers $q_1$ and $m_1$ as well.
Then, $f(s/t)=f(1/t+...+1/t)=f(1/t)^s=(f(1)^{1/t})^s=f(1)^{s/t}$, for any rational number.
Since the rationals are dense in the reals and $f$ is continuous, for any real number $x$ we can find a sequence $(r_n)$ that will converge to $x$ with $f(r_n)=f(1)^{r_n}$, so that $f(x)=f(1)^x$ for all real numbers.
Since $f(1)>0$ and is real-valued, $\log(f(1))=c$ exists.  Then:
$f(x)=f(1)^x=(\exp(\log(f(1))))^x=(\exp(c))^x=(e^c)^x=e^{cx}$.
Thanks again!
A: $f(x)= f(\frac{x}{2})f(\frac{x}{2}) $ and together with $f(x) \neq 0$ implies $f(x)$ is positive (actually if  $f(x) $ is zero anywhere then it is zero everywhere)
Define $g(x)= log f(x)$
Then $g(x) + g(y) =g(x+y)$ which is cauchy functional equation with solution $g(x) = cx$ where $c$ is a constant
So  $f(x)= e^{cx}$
A: First note that $f(x) > 0$, for all $x \in \mathbb{R}$. This can be seen from the fact that
$$f(x) = f\left(\dfrac{x}2 + \dfrac{x}2\right) = f \left(\dfrac{x}2\right)^2$$ Further, you can eliminate the case $f(x) = 0$, since this would mean $f \equiv 0$.
One way to go about is as follows.
$1$. Prove that $f(m) = f(1)^m$ for $m \in \mathbb{Z}^+$.
$2$. Now prove that $f(m) = f(1)^m$ for $m \in \mathbb{Z}$.
$3$. Now prove that $f(p/q) = f(1)^{p/q}$ for $p \in \mathbb{Z}$ and $q \in \mathbb{Z} \backslash \{0\}$.
$4$. Now make use of the fact that rationals are dense in $\mathbb{R}$ and hence you can find a sequence of rationals $r_n \in \mathbb{R}$ such that $r_n \to r \in \mathbb{R} \backslash \mathbb{Q}$. Now use continuity to conclude that $f(x) = f(1)^x$ for all $x \in \mathbb{R}$. You will see that you need only continuity at one point to conclude that $f(x) = f(1)^x$.
