BehindEnemyLines
Senior member
Here's a problem I've been doing for some time already:
The half-life of the odor left behind by the dog is 40 seconds. How much of the odor remains after 3 minutes (180 seconds)?
This is what I got:
The equation for exponential decay and half-life is A = Az * e^kt, k < 0 & Az is the initial amount. Since only half-of Az remains after 40 secs, then:
(1/2)Az = Az*e^40k
1/2 = e^40k
ln(1/2) = ln(e^40k)
ln(1)-ln(2) = 40k
k = -ln(2)/40
So, A = Az*e^(-t*ln(2)/40) or A = Az(1/2)^t/40. However, since there's no specified initial amount Az, how do I find the amount after 180 seconds? May be the teacher wants just the equation, like so:
A = Az*e^(1/2)^(180/40), where Az can be any initial amount.
The half-life of the odor left behind by the dog is 40 seconds. How much of the odor remains after 3 minutes (180 seconds)?
This is what I got:
The equation for exponential decay and half-life is A = Az * e^kt, k < 0 & Az is the initial amount. Since only half-of Az remains after 40 secs, then:
(1/2)Az = Az*e^40k
1/2 = e^40k
ln(1/2) = ln(e^40k)
ln(1)-ln(2) = 40k
k = -ln(2)/40
So, A = Az*e^(-t*ln(2)/40) or A = Az(1/2)^t/40. However, since there's no specified initial amount Az, how do I find the amount after 180 seconds? May be the teacher wants just the equation, like so:
A = Az*e^(1/2)^(180/40), where Az can be any initial amount.
