Answer:
Given
[tex]f(m+n)=f(m)f(n)[/tex]
If we assume
[tex]f(x)=ae^{x}\\\\f(m+n)=ae^{m+n}\\\\\therefore f(m+n)=ae^{m}\times ae^{n}(\because x^{a+b}=x^{a}\times x^{b})\\\\\Rightarrow f(m+n)=f(m)\times f(n)[/tex]
Similarly
We can generalise the result for
[tex]f(x)=am^{x }[/tex] where a,m are real numbers
2)
[tex]f(m\cdot n)=f(m)+f(n)\\\\let\\f(x)=log(x)\\\therefore f(mn)=log(mn)=log(m)+log(n)\\\\\therefore f(mn)=f(m)+f(n)[/tex]
