If the number N of bacteria present in a culture at time t (in hours) obeys the function?

Question

N (t) = 1000e0.01t. After how many hours will the population equal 1500? 2000?

Also, if Iodine 131 is a radioactive material that decays according to the function
A(t) = A0 e-0.0244t where A0 is the initial amount present and A is the amount present at the time t (in days).
a)What is the half-life of iodine 131?



b) Determine how long it takes for 100 grams of iodine 131 to decay to 10 grams.

I REALLY NEED HELP WITH LOGS. I don't understand them at all.

Answer 1

to solve this problem, you must first understand the equation. The general formula is N(t) = N* e^(-RT) , where

N(t) = amount at the moment / you need at final point
N = initial amount
R = is a constant, which is different from problem to problem
T = is the time, in seconds, minutes, years, etc

Now that we have that down.... After how many hours will the population equal 1500 and 2000?

Well, N(t) = 1500, so we have...

1500 = 1000 * e^(-.01t) , taking ln of both sides we get:

ln(1500) = -.01t * ln (1000*e) this implies:

[ ln (1500) ] / [ -.01 * ln (1000e) ] = t

same thing for 2000.

half goes as follows:

If you are looking for the HALF LIFE then you need the final amount to be 1/2 of the initial, or

A final = 1/2 A0. this translates into

.5 A0 = A0 * e^(-.0244t), using same idea as above, solve for it.

b) should be easy to do now, A final = 10 grams, A0 = 100 grams, you know the R value (-.0244), just plug it into the generic formula, take logs of both sides as I showed above and you should be good!

Hope this was helpful. Good luck

Answer 2

[1] ... N(t) = 1000 e^(0.01 t)

We want to solve

[2] ... N(t) = 1500
[3] ... 1000 e^(0.01 t) = 1500
[4] ... e^(0.01 t) = 1.5

In order to "undo" the power of e, take Ln

[5] ... 0.01 t = ln 1.5 = 0.4055
[6] ... t = 40.55

For N(t) = 2000, follow same method; you will find t = 100 * ln 2 = 69.31

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[7] ... A(t) = A0 * e^(-0.0244 t)

We want to know after how much time half of A0 is left, that is

[8] ... A(t) = 0.5 * A0
[9] ... A0 * e^(-0.0244 t) = 0.5 * A0
[10] ... e^(-0.0244 t) = 0.5

Take the Ln again,

[11] ... -0.0244 t = ln 0.5 = -0.6931
[12] ... t = (-0.6931) / (-0.0244) = 28.41

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For the last part, you know A0 = 100.

[13] ... 100 * e^(-0.0244 t) = 10
[14] ... e^(-0.0244 t) = 0.1
[15] ... -0.0244 t = ln 0.1 = -2.3026
[16] ... t = (-2.3026) / (-0.0244) = 94.37

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For this kind of problems, remeber: LOG is the opposite of 10^X; and LN is the opposite of e^x.

So any equation of the form

e^x = a

can be rewritten as

x = ln a

etc.