∵f(x+y)=f(x)f(y),∀x,y∈R....... (i). Put x=y=1, we get f(2)=f(1).f(1)=9[∵f(2)=9] ⇒f(1)2=9⇒f(1)=3 Now, put x=2 and y=1 in Eq. (i), we get f(3)=f(2).f(1)=32.3=33 Now, put x=3 and y=1 in Eq. (i), we get f(4)=f(3).f(1)=33.3=34 Again, put x=4 and y=2 in Eq. (i), we get f(6)=f(4).f(2)=34.32=36