Question

Let $f$ be a continuous function on R satisfying $f(x+y)=f\left(x\right)f\left(y\right)$ for all $x,y\in R$ and $f\left(1\right)=4$ then $f\left(3\right)$ is equal to

Answer 1

Substituting $x=y=0$, we obtain $f\left(0\right)=\left(f\right(0){)}^2$ so $f\left(0\right)=0$ or $f\left(0\right)=1$. If $f\left(0\right)=0$ then $f\left(0\right)=f(1+0)=f\left(1\right)f\left(0\right)=0$ not true.
Hence $f\left(0\right)=1$. Replacing $y$ by $x/2$ and $x$ by $x/2$, we have $f\left(x\right)=\left(f\right(x/2){)}^2>0$
for all x. $1=f\left(0\right)=f\left(x\right)f(-x)$ so $f\left(x\right)\ne 0$ for all $x$).
Now define
$g\left(x\right)=\log f\left(x\right)$ so $g(x+y)=g\left(x\right)+g\left(y\right)$ and $g$ is continuous so
$g\left(x\right)=g\left(1\right)x=\left(\log f\right(1\left)\right)x\Rightarrow f\left(x\right)=e^{kx}=a^x$
where $a=e^k=e^{\log f\left(1\right)}=f\left(1\right)=4$. Thus
$f\left(x\right)=4^x\Rightarrow f\left(3\right)=4^3=64$