Word problem concerning logs?

[imagesparkle]

The half life of a radioactive substance is 30 days. If 20 grams are initially present, how much time has passed if only 3 grams remain?

I know that you have to use the formula: A=(a)e^k(t) but how do I go about it?

 

[JohnZ]

A=(a)e^kt.
use initial condition to determine (a).
use half lift to determine k
then solve the problem.

 

[HallsofIvy]

The half life of a radioactive substance is 30 days. If 20 grams are initially present, how much time has passed if only 3 grams remain?

I know that you have to use the formula: A=(a)e^k(t) but how do I go about it?

Have you been able to solve this using JohnZ's hints?

"20 grams are initially present"- so when t= 0, A= 20. Putting those into the formula, 20= a(e^0). Solve that for a. Do you know what e^0 is?

"The half life of a radioactive substance is 30 days." Do you know what "half life" means? It is the time until only half of the initial amount is left- if you start with A= 20 g, when t= 30, A will be 20/2= 10. Putting that into the formula, 10= 20e^(30k). Solve that for k. You will need to use a natural logarithm to get rid of that exponential.

To determine how much time has passed if only 3 gram remain, solve 3= Ae^(kt) for t with A and k equal to the values you have found.