Question

The number of bacteria present in a culture at any time , t hours, can be modelled by the equation N=Noe^kt
If the original number is doubled in 3 hours, find k

Answer 1

remember if the time t=0, the value will be No
\[\large N(0)=N _{0}*e^0=N _{0}\]

Answer 2

They tell you the number has doubled at 3 hours, You will have 2*No, when t=3

\[\large N(3)=2*N _{0}=N _{0}*e^{3t}\]

Answer 3

sorry it is the k left in the exponent.. put in t=3
\[\large 2*N _{0}=N _{0}*e^{3k}\]

Answer 4

Solve for the k value,

you get all that?

Answer 5

a little lost @DanJS

Answer 6

ok. The exponential form they give you ,
\[\large N=N _{0}*e^{kt}\]
Shows the number bacteria N, for any time t.
The No is the original number bacteria at the start, when t=0.
Putting in t=0 will show you,
\[N=N _{0}*e^0 = N _{0}*1 = N _{0}\]

you get that far?

Answer 7

N=No*e^kt
Where No is beginning amount and N is ending amount
where "e" is the mathematical constant 2.718281828
If No is (let's say 100 for example)
Then after 3 hours "N" will be 200
So, the equation then becomes
200 = 100 * e^k*3

Answer 8

okay i am good so far with (0, No) .. if i am not mistaken

Answer 9

Then it says, the original number was doubled in 3 hours.

The original number is No, from when t=0. So double that is 2*No, when t=3

Answer 10

\[\large N=N _{0}*e^{k*t}\]

For t=3, you have double original number 2*No
So if you put t=3 into that equation, you get 2*No

\[\huge 2*N _{0}=N _{0}*e^{k*3}\]

Answer 11

that you can solve for k, divide both sides by No first
\[\large 2=e^{3k}\]

Take natural log of both sides
\[\large \ln(2)=\ln e^{3k}\]

Answer 12

recall the property for natural log base e of e gives
\[\large \ln e^a =a\]

so you have
\[\large \ln(2)=3k\]
\[\large k=\frac{ \ln(2) }{ 3 }\]

Answer 13

oh yup! got it :D Thank you so much

Answer 14

welcome