SOLUTION: 10. The half-life of radioactive potassium is 1.3 billion years. If 10 grams is present now, how much will be present in 100 years? In 1000 years? 12.A fossilized leaf contains

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: 10. The half-life of radioactive potassium is 1.3 billion years. If 10 grams is present now, how much will be present in 100 years? In 1000 years? 12.A fossilized leaf contains       Log On


   



Question 65624This question is from textbook
: 10. The half-life of radioactive potassium is 1.3 billion years. If 10 grams is present now, how much will be present in 100 years? In 1000 years?
12.A fossilized leaf contains 70% of its normal amount of carbon 14. How old is the fossil?
This question is from textbook

Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
A = Pe^(rt)
P/2 = Pe^(1,300,000,000r)
1/2 = e^(1,300,000,000r)
ln(0.5) = 1,300,000,000r
ln(0.5)/1,300,000,000 = r
r is about -5.33e-10
A = 10e^(r*100)
and
A = 10e^(r*1000)
I do not have a calculator with "e" right now.
0.7P = Pe^(rt)
0.7 = e^(rt)
I need to know the rate ....