Exponential and logarithmic models

Question

The growth count in a bacteria culture was 400 after 10 minutes and 1300 after 30 minutes. Assuming the count grows exponentially,

what was the initial size of the culture? ____

find the doubling period. _____

find the population after 80 minutes.____

when will the population reach 11,000? ____

you may enter exact value or round up to 2 decimal places.

Answer 1

400=pe^10r

1300=pe^30r

400/p = e^10r

ln(400/p) =10r

r= ln(400/p)/10

1300/p =e^30r

ln(1300/p)=30r

r=ln(1300/p)/30

set the two values of r equal:

ln(400/p)/10=ln(1300/p)/30

3ln(400/p) = ln(1300/p)

ln(1300/p)/ln(400/p) = 3

p=800/sqr13 = about 221.88= culture's initial size

doubling time = about 11 minutes and 46 seconds

in 80 minutes

b = (800/sqr13)e^80r

where r = ln((sqr13)/2)/10= about .05893275= about 5.9% per minute

b=about (800/sqr13)e^80(.059)

=about 24,754.35

it double about every 12 minutes

1300 in 30 minutes

2,600 in about 42 minutes

5,200 in about 54

10,400 in about 66

20,800 in about 78 minutes

11,000 in a little over 66 minutes

11,000 =221.88e^.059t

11000/221 = e^.06t

t = ln11000/221)/.06

t=about 66.31 minutes = about 66 minutes and 18 seconds

Answer 2

400 = a₀e^10k

1300 = a₀e^30k

3.25 = e^20k

ln 3.25 = 20k

Solve for k

Substitute into either origianl equation to determone a₀

Substitute k and a₀ into 2 = a₀e^kt and solve for 2

A(80) = a₀e^k(80), solve for A(80)

11000 = a₀e^kt, solve for t