Question 134079: Find the half-life of a radioactive substance that is reduced by 30% in 20 hours
Answer by Madgamer107(9)   (Show Source):
You can put this solution on YOUR website! this problem can be most easily solved by using the Pe^kt formula. Because all that is given is a 30% increase in 20 hours, you can assign values to the final amount and the initial. For example...
70 = 100e^(k x 20)
This formula works because there is a 30% reduction (from a principle value of 100 to 70) and the rate is designated 20. Solve the the constant, k.
70 = 100e^(20k)
.7 = e^(20k)
ln .7 = 20k ln e
ln .7 = 20k
(ln .7)/20 = k
typing this into a calculator, you get a k of -.0178. Now you have to solve for the half-life, or the amount of time it wil take to decompose 50%. Make a new equation (bear in mind you now know k), and solve for r, the amount of time it will take.
50 = 100e^(-.0178t)
.5 = e^(-.0178t)
ln .5 = -.0178t ln e
ln .5 = -.0178t
(ln .5) / (-.0178) = t
Keying that into a calculator will give you a time of 38.941 hours. Finally, make sure this is a feasible answer. The problem stated that it takes 20 hours to decompose 30%, so this answer does make sense.
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