Question

Help! I dont know what equation to use for this to find my initial population
The count in a bateria culture was 400 after 20 minutes and 1100 after 40 minutes.
What was the initial size of the culture?
Find the doubling period.
Find the population after 95 minutes.
When will the population reach 10000?

Answer 1

so is this exponential growth?

Answer 2

yes

Answer 3

or is there a growth model?

Answer 4

im missing two variables from my exponetial growth formula that i have when trying to complete this question

Answer 5

\[P = P_{0} e^{kt}\]

Answer 6

how do you know what K is though?

Answer 7

well lets start with the initial population

you know

\[400 = P_{0} e^{20k}\]

\[1100 = P_{0}e^{40k}\]

Answer 8

I dont see how that helps me, when i'm still missing two variables. this is where i dont know where to go from

Answer 9

hey...do we assume that bacteria goes binary fission

Answer 10

ok so

\[P_{0} = \frac{400}{e^{20k}}\]

and in the 2nd equation \[P_{0} = \frac{1100}{e^{40k}}\]

now you can equate the 2

Answer 11

because if we do..it'll be easy..

Answer 12

and solve fof the growth constant k...

when you have that, let t = 0 to find the initial population

Answer 13

so you put 400/(e^20k)=11/(e^40k) ? yes?

Answer 14

then we can have....the population growth growing like x,2x,4x....each after time t...if t be the time for one one fission...and x be initial population...
then number of times fission occurs is(20/t) for the first tym, and (40/t) for the second condition...now in these two cases since we have exponential growth...the population will be..2^(20/t)*x=400 and 2^(40/t)*x=1100...now solve simultaneously....t comes as 5.78...is it correct..?

Answer 15

my opinion note the time with x and the number of population with y
so and us the rule of 3 simple

y=400 when x=20
y=1100 when x=40

1. y=? when x=0 the first question
2. when y=2y so than x=?
3.when x=95 so than y=?
4.y=10000 when x=?

can you calcule it from these ?

Answer 16

thats correct @darkmare

so

\[\frac {400}{1100} = \frac{e^{20k}}{e^{40k}}\]

then using log laws

\[\frac{4}{11} = e^{-20k}\]

take the log of both sides

\[\ln(\frac{4}{11}) = -20k\]

\[k = \frac{-1}{20} \times \ln(\frac{4}{11})\]


then k = 0.05058 the growth constant/

so the model is now

\[P = P_{0} e^{0.05058t}\]

Answer 17

next you need to find the initial population so using one of the known equations (400 and 20 mins)

\[400 = P_{0} \times e^{0.05058 \times 20}\]

then

\[P_{0} = \frac{400}{e^{0.05058 \times 20}}\]

just evaluation to find the initial population.

now you have your model.

Answer 18

Thank you!! That made a lot of sense after you typed it out for me!