The count in a bacteria culture was 200 after 10 minutes and 1100 after 35 minutes. Assuming the count grows exponentially,

Question

What was the initial size of the culture?   

Find the doubling period.   

Find the population after 60 minutes.   

When will the population reach 11,000?

Answer 1

You might find this way easier

A(t)=A*b^t,,,A=A(0), and b is the base of exponentiation (b>1 means exponential growth)

for this problem

200=A*b¹⁰

1100=A*b³⁵

b³⁵/b¹⁰=b²⁵=(1100/200)

b=(1100/200)^(1/25)

b=1.0706 approx.

calculating A

200=A(1.0706)¹⁰

A=101

Amount at t equation is

A(t)=101*1.0706^t

notice that for each minute increase

amount increases 107.06%

say A(t₂)=2*A(t₁)

then 2=1.0706^(t2^-t1⁾

ln(2)=(t₂-t₁)*ln(1.0706)

t₂-t₁=10.16 approx. minutes

if t=60

A=101*(1.0706)⁶⁰

A=6053 appox.

if A(t)=11000

11000=101*(1.0706)^t

t=[ln(11000/101)]/[ln(1.0706)

t=68.75 approx. minutes

Answer 2

A = Ke^rt

200 = Ke^10r

1100 = Ke^35r

1100/200 = e^35r/e^10r = e^25r = 5.5

25r = ln(5.5) --> r = ln(5.5)/25

A = Ke^ln(5.5)t/25 = K(5.5)^t/25

1100 = K(5.5)^35/25 = K(5.5)^1.4 = 10.88K

K = 1100/10.88 ≈ 101

1) At t = 0 , A = 101

2) 202 = 101(5.5)^t/25

ln(2) = (t/25)ln(5.5)

t = 25ln(2)/ln(5.5) ≈ 10.165 minutes

3) A = 101(5.5)^60/25 = 101(5.5)^2.4 ≈ 6042

4) 11000 = 101(5.5)^t/25

t/25 = ln(108.91)/ln(5.5)

t = 25(2.75) ≈ 68.786 minutes