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Mathematics 15 Online
OpenStudy (anonymous):

A culture started with 1,000 bacteria. After 6 hours, it grew to 1,000 bacteria. Predict how many bacteria will be present after 15 hours. Round your answer to the nearest whole number. P=Ae^kt

OpenStudy (anonymous):

some typo here yes?

OpenStudy (anonymous):

start with 1,000 grow to 1,000 means it did not grow at all.

OpenStudy (sriram):

i agree

OpenStudy (anonymous):

Sowwy :( it grew to 1,100

OpenStudy (sriram):

P=Ae^kt what does this signify?

OpenStudy (anonymous):

ok it grew by 10% yes?

OpenStudy (anonymous):

The formula. & yes

OpenStudy (anonymous):

we have a choice. we can use the stupid formula you wrote, or we can do it an easier way. which do you choose?

OpenStudy (sriram):

okay A=1000 t=6 hrs P=1100 1100=1000* e^6k e^6k=1.1 solve for k

OpenStudy (anonymous):

i will let sriram use the formula if desired. but we can do this. it increases by 10% every 6 hours so i would just write \[A=1000\times(1.1)^{\frac{t}{6}}\]

OpenStudy (sriram):

using this value of k P=1000*e^15k solve for P

OpenStudy (anonymous):

then i replace t by 15 and i am done. i get \[A=1000\times (1.1)^{\frac{15}{6}}=1269\] finished

OpenStudy (anonymous):

we can work out the annoying \[P=P_0e^{kt}\] if you like but it takes several steps. first note that when t = 6 you know P = 1100 so write \[1100=1000e^{6k}\] divide by 1000 to get \[1.1=e^{6k}\] now solve for k to get \[\ln(1.1)=6k\] \[k=\frac{\ln(1.1)}{6}\]

OpenStudy (anonymous):

now use a calculator to get \[k=0.0158850\]

OpenStudy (anonymous):

now go back and write \[P=1000e^{.0155885t}\] and finally replace t by 15 to get \[P=1000e^{.05885\times 15}\] \[=1000e^{0.238275}\] \[=1269\]

OpenStudy (anonymous):

Thank youuuu <3

OpenStudy (anonymous):

same answer except first way takes like two seconds and second way is a pain in the log

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