SOLUTION: What is the half-life of a material that decays exponentially at the rate of 3.54% per year? Using exponential decay formula N(t)= Noe^(-kt) ? Thank you

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: What is the half-life of a material that decays exponentially at the rate of 3.54% per year? Using exponential decay formula N(t)= Noe^(-kt) ? Thank you      Log On


   

Question 1118557: What is the half-life of a material that decays exponentially at the rate of 3.54% per year? Using exponential decay formula N(t)= Noe^(-kt) ? Thank you
Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39621) About Me  (Show Source):
You can put this solution on YOUR website!
One year of time pass
1%2Ae%5E%28-k%2A1%29=1-0.0354

e%5E%28-k%29=0.9646

ln%28e%5E%28-k%29%29=ln%280.9646%29

-k=-0.03604

k=0.03604
-

The model equation: highlight_green%28N%28t%29=N%5Bo%5De%5E%28-0.03604t%29%29
FINDING HALF-LIFE
e%5E%28-0.03604t%29=0.5%2F1
e%5E%28-0.03604t%29=0.5
ln%28e%5E%28-0.03604t%29%29=ln%280.5%29

t=ln%280.5%29%2F%28-0.03604%29

t=highlight%2819.23%29 or about 19 years 3 months

Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


Scientists like to use the A+=+Pe%5E%28-kt%29 formula for working problems involving half lives. To me that is a lot of extra work, because you have to use the given information to find the k value and then use the k value to find the half life.

It seems far more straightforward to simply ask how many times does the original amount need to be reduced by the decay factor to end up with half as much as you started with.

For your problem, with the given information that the decay rate is 3.54% per year, we see that 96.46% of the original amount REMAINS after a year, so the decay factor is 0.9646. Then we only need to solve

0.9646%5Ex+=+0.5
x%2Alog%28%280.9646%29%29+=+log%28%280.5%29%29
x+=+log%28%280.5%29%29%2Flog%28%280.9646%29%29+=+19.23

The half life is 19.23 years.