Question

At a certain time, bacteria in a culture has 115,200 bacteria. Four hours later, the culture has 1,440,000 bacteria. How many bacteria will there be in the culture in 15 hours?

Answer 1

I got: `1,495,770,721` BUT check my maths!

Explanation:

I would call the number of bacteria `y` and write the exponential function:
`y=Ae^(kt)`
with `t` the time and `A and k` two constants.

With your numerical data we have:

At `t=t_1`

`115,200=Ae^(kt_1)`

At `t=t_2=t_1+4` (4 hours later):

`1,440,000=Ae^(k(t_1+4))`
that can be written as:
`1,440,000=Ae^(kt_1)e^(4k)`

we can substitute the first equation: `115,200=Ae^(kt_1)` into the second to get:
`1,440,000=color(red)(115,200)e^(4k)`
solve it for `k`:
`e^(4k)=(1,440,000)/(115,200)=12.5`
using natural logs on both sides we get:
`4k=ln(12.5)`
`k=1/4ln(12.5)=0.63143`

At `t=t_3=t_1+15` (15 hours later)

`y=Ae^(k(t_1+15))`
that can be written as:
`y=Ae^(kt_1)e^(15k)`
again we substitute the first equation and the value of `k` found previously:
`y=115,200*e^(15*0.63143)=1,495,770,721`

Answer 2

`1,495,770,721` bacteria will be in `15` hours.

Explanation:

Formula for exponential growth is
`y(t) = a * e^(kt)`

Where y(t) = Number at time "t" , a = Number at start.
k = rate of growth (when >0) or decay (when <0) ,t = time

`1440000 = 115200 * e^(k*4) or e^(4k)=1440000/115200=12.5` Taking log on both sides we get `k*4= ln(12.5) or k = ln(12.5)/4= 0.6314216`

After `15` hrs bacteria will grow to `y(15) =115200 * e^(k*15) = 115200 * e^(0.6314216 *15) = 1495770721`

`1,495,770,721` bacteria will be in `15` hours. [Ans]