Question

Ifthegrowthrateofthenumber of bacteria at any time t is proportional to the number present at t and doubles in 1 week, how many bacteria can be expected after 2 weeks? After 4 weeks?

Answer 1

et x be the number of bacteria. The growth rate is the change in x over time dxdt
growth rate proportional to the number present means
dxdt=k⋅x

where k is the constant of proportionality

Answer 2

oops

Answer 3

i was able to get to that part \[\frac{ dx }{ dt }=k*x\]

Answer 4

\[\large
\frac{dx}{x} = k*dt \\ \large
\ln(x) = kt \\ \large
x = x_0e^{kt}
\]

Answer 5

where \(x_0\) is the number of bacteria at time t = 0.
When t = 1, \(x = 2x_0\): \(2x_0 = x_0e^k;~~~~2 = e^k; ~~~~k = \ln(2)\)

Answer 6

so then id place 2x0=e^ln(2)?

Answer 7

\[
x = x_0e^{kt} \\
k = \ln(2) \\
x = x_0e^{\ln(2) * t} = x_0(e^{\ln(2)})^t = x_02^t \\
\text{When t = 2, }x = x_02^2 = 4x_0 ~~\text{ (4 times the initial population in 2 weeks.) } \\
\text{When t = 4, }x = x_02^4 = 16x_0 ~~\text{ (16 times the initial population in 4 weeks.) } \\
\]

Answer 8

ohhh okay!

Answer 9

\(e^{\ln(2)} = 2 \)

Answer 10

sweet thanks aum!