A bacterial Culuture starts with 500 bacteria and after 10 m

Solution

Let
P(t) the population at time t
We have
P(t)=C*e^kt

When t=0, P(0)=C=500

So we have P(t)=500e^kt

when t=10, P(10)=500e^k*10=600, so e^k*10=600/500=6/5

ln(e^k*10)=ln(6/5)

10k=ln(6/5), so k=ln(6/5)/10

To find the time when there are 500 bacteria

P(u)=500e^ku=5000

e^ku=5000/500=10

ku=ln(10)

u=ln(10)/k=ln(10)/(ln(6/5)/10)=10ln(10)/ln(6/5)126.29