SOLUTION: the half life of radioactive potassium is 1.3 billion years. If ten grams are present now, how much will be present in 100 years? 1000 years?

Question 242015: the half life of radioactive potassium is 1.3 billion years. If ten grams are present now, how much will be present in 100 years? 1000 years?
Answer by nyc_function(2741) (Show Source): You can put this solution on YOUR website!
We use M(x) = c(.5^(x/h)), where c = initial mass, h = half-life and x = years.
NOTE: M(x) = the total amount of grams after certain years.
We first need to solve for h.
1.3 = 10(.5^(100/h))
1.3/10 = .5^(100/h)
0.13 = .5^(100/h)
Take log of both sides.
ln(0.13) = ln(.5^(100/h)
ln(0.13) = (100/h)(ln(.5))
Multiply both sides by h.
h(ln(0.13)) = 100(ln(.5))
Divide both sides by ln(0.13) to find h.
h ≈ 100(ln(.5))/ln(0.13)
h ≈ 33.97412529
We can round that off to the nearest year and the decimal number becomes 34 as the half-life.
I did enough. All you have to do now is replace x with 100 and simplify and then do the same but using x = 1000 in the formula and simplify.
Can you take it from here?

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