In a certain culture of bacteria, the rate of increase is proportional to the number present. If there are 10^4 at the end of 3 hours and 4 10^4 at the end of 5 hours, then there were in the beginning.

Step-by-step solution

Let x be the number of bacteria present at time t . array ll & d x dt x & ~d x dt = k x & ~d x x = kdt array Integrating on both sides, we get x= kt + c ...(i) When t =3, x=10^4=10,000 Equation (i) becomes (10,000)=3 k + c ...(ii) When t=5, x=4.10^4=40,000 Equation (i) becomes (40,000)=5 k + c ...(iii) Subtracting (ii) from (iii), we get k= 2 From equation (ii), (10 ; 000)=3 2+c c= ( 10^4 8 ) Now, Initially t =0 From (i), aligned & x= k 0+ ( 10^4 8 ) & x= ( 10^4 8 ) & x= 10^4 8 aligned