Exponential Growth of Bacteria

[τheory]

Homework Statement

A bacteria culture starts with 840 bacteria and grows at a rate proportional to its size. After 3 hours there will be 2520 bacteria.

How long will it take for the population to reach 1030 ?
2. The attempt at a solution

dP/dt = k*P
dP/P = k*dt

*Integrated both sides
ln |P| = k*t + C
P = e^(k*t + C)

*Used (0, 840) as initial values
840 = e^(k*0 + C)
C = ln840

P = e^(k*t + ln840)
P = 840e^(k*t)

*Used (3, 2520) to solve for k
P = 840e^(k*t)
2520 = 840e^(k*3)
k = 0.366

*Solved t when P = 1030
P = 840e^(0.366*t)
1020 = 840e^(0.366*t)
t = 0.557

I submit this as the answer on Webwork but it says the answer is wrong; can anyone explain to me why?

 

[Ray Vickson]

Homework Statement

A bacteria culture starts with 840 bacteria and grows at a rate proportional to its size. After 3 hours there will be 2520 bacteria.

How long will it take for the population to reach 1030 ?
2. The attempt at a solution

dP/dt = k*P
dP/P = k*dt

*Integrated both sides
ln |P| = k*t + C
P = e^(k*t + C)

*Used (0, 840) as initial values
840 = e^(k*0 + C)
C = ln840

P = e^(k*t + ln840)
P = 840e^(k*t)

*Used (3, 2520) to solve for k
P = 840e^(k*t)
2520 = 840e^(k*3)
k = 0.366

*Solved t when P = 1030
P = 840e^(0.366*t)
1020 = 840e^(0.366*t)
t = 0.557

I submit this as the answer on Webwork but it says the answer is wrong; can anyone explain to me why?

A more precise answer is t = 0.5568266199 hr. Perhaps the Webwork system wants more accuracy than you provided; in this Forum we have seen many instances where on-line answers have been rejected because of insufficient numbers of digits. You could try re-submitting a more accurate value, such as 0.5568 or 0.55683. Or, maybe Webwork does not want the '0' before the decimal, so might prefer you to write .557 or .5568. (Actually, that sounds like a badly-written system, but what can you do?)

BTW: in computations of this type, ALWAYS use more digits _during_ the computation, possibly rounding off the final answer; that helps to reduce the chance of incorrect results due to roundoff accumulation. For example, you should use a more accurate value of k; the precise value is k = (1/3)*ln(3) =~= 0.3662040963 =~= 0.36620.

RGV

 

[τheory]

Thanks, I tried resubmitting with the following:
$\frac{3ln(103/84)}{ln3}$

Which equates to about 0.556 and gave me the right answer.