Let x be the number of bacteria at any time t. Given that,
dx
dt
∝x(∵ Rate of growth =
dx
dt
) ⇒
dx
dt
=λx⇒
dx
x
=λdt After integrating it, we get logx=λt+C . . . (i) =0,x=1000 which gives log1000=0+C⇒C=log1000 Given, when t=0,x=log1000 logx−log1000=λt or log(
x
1000
)=λt . . . (ii) Given that in 2h, number of bacteria increased by 20% i.e. when t=2h,x=1200 Put, t=2 and x=1200 in Eq. (ii), log(
1200
1000
)=2λ gives, λ=
1
2
log(
6
5
) Again, from Eq. (ii), log(
x
1000
)=
1
2
log(
6
5
) or
x
1000
=e
t
2
log(
6
5
). . . (iii) Given, x=2000 at t=k∕loge(6∕5), put in Eq. (iii),