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Mathematics 25 Online
OpenStudy (anonymous):

the number N of bacteria in a culture is modeled by N=250e^kt, where t is the time in hours. if N=280 when t=10, estimate the time requires for the population to double in size.

OpenStudy (lgbasallote):

i suggest you first find what is k

OpenStudy (lgbasallote):

can you do that?

OpenStudy (anonymous):

i tried and i get 0.11 but im nto sure if its right

OpenStudy (anonymous):

when i plugged it back in tot he equaton again it turned out to be 7.33 hours, but thats not the right answer D:

OpenStudy (lgbasallote):

you have \[\ln N = kt\] right..so k is \[\frac{\ln N}{t} = k\] is this what you did?

OpenStudy (lgbasallote):

wait...

OpenStudy (lgbasallote):

divide both sides by 250 first... \[\frac{N}{250} = e^{kt}\] then take ln \[\ln (\frac{N}{250}) = kt\] divide by t \[\frac{\ln (\frac{N}{250})}{t} = k\] is this what you did?

OpenStudy (anonymous):

i thoguht it was 280=250e^k(10)?

OpenStudy (lgbasallote):

yup it is...

OpenStudy (lgbasallote):

that's also right...but you will still end up doing what i did

OpenStudy (lgbasallote):

\[280 = 250e^{10k}\] \[\frac{280}{250} = e^{10k}\] \[\ln (\frac{280}{250}) = 10k\] \[\frac{\ln(\frac{280}{250})}{10} = k\] see?

OpenStudy (lgbasallote):

so is that what you did to find k?

OpenStudy (anonymous):

yep

OpenStudy (anonymous):

i got 0.11

OpenStudy (lgbasallote):

not 0.011?

OpenStudy (anonymous):

uh oh... that's what happened!! oh gosh ahah thank you

OpenStudy (anonymous):

no wonder the answer came out wrong

OpenStudy (lgbasallote):

\[\ln (\frac{280}{250}) = 0.11\] you still need to divide by 10

OpenStudy (anonymous):

wait so k isnt =0.011?

OpenStudy (lgbasallote):

k is 0.011

OpenStudy (anonymous):

and then my new equation is 560=250e^0.011t right?

OpenStudy (lgbasallote):

would you like to know a quick formula on how to find the time when this thing will double?

OpenStudy (anonymous):

yes please

OpenStudy (lgbasallote):

use this formula \[\frac{\ln 2}{k}\]

OpenStudy (lgbasallote):

would you like to know the derivatin behind this?

OpenStudy (anonymous):

yes

OpenStudy (lgbasallote):

okay..the formula for exponential growth is \[\ln (\frac{x}{x_o} )= kt\] x is the final population x_0 is the initial population t is the time k is the constant so let's call the time when population will double as t" (notice the double-apostrophe) so at t", final population will be twice the initial so \[x = 2x_o\] if i substitute this into the formula for exponential growth \[\ln(\frac{2x_o}{x_o}) = kt"\] \[\ln(\frac{2\cancel{x_o}}{\cancel{x_o}}) = kt"\] \[\ln 2 = kt\prime\] remember we want t' so we divide both sides by k \[t "= \frac{\ln 2}{k}\]

OpenStudy (lgbasallote):

that's supposed to be \[\ln 2 = kt"\]

OpenStudy (lgbasallote):

does that make sense?

OpenStudy (anonymous):

you are good at explaining, thank you :D

OpenStudy (lgbasallote):

haha thanks ^_^

OpenStudy (lgbasallote):

i can also do it from where you were going a while ago since k = 0.011 the new equation would be 500 = 250e^0.011t see how 500 is twice 250? what you did wrong was what you multiplied by 2 was the population at time = 10 should've been the 250 this one will give you the same answer :)

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