SOLUTION: The half-life for thorium-227 is 18.72 days. The amount A (in grams) of thorium-227 after t days for a 10-gram sample is given by A(t)=10⋅0.5^t/18.72 How long will it take bef

Algebra ->  Exponents -> SOLUTION: The half-life for thorium-227 is 18.72 days. The amount A (in grams) of thorium-227 after t days for a 10-gram sample is given by A(t)=10⋅0.5^t/18.72 How long will it take bef      Log On


   

Question 1156050: The half-life for thorium-227 is 18.72 days. The amount A (in grams) of thorium-227 after t days for a 10-gram sample is given by A(t)=10⋅0.5^t/18.72
How long will it take before 3 grams of thorium-227 is left in the sample? Round your answer to the hundredths place.
....days
Thank you
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


3+=+10%280.5%5E%28t%2F18.72%29%29
0.3+=+0.5%5E%28t%2F18.72%29
log%28%280.3%29%29+=+%28t%2F18.72%29%2Alog%28%280.5%29%29
t%2F18.72+=+log%28%280.3%29%29%2Flog%28%280.5%29%29
t+=+18.72%2A%28log%28%280.3%29%29%2Flog%28%280.5%29%29%29 = 32.515995..., or 32.52 to the nearest hundredth.

ANSWER: 32.52 days

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Added comment....

I find that using the given formula A%28t%29+=+10%280.5%5E%28t%2F18.72%29%29 to solve the problem adds an unnecessary level of difficulty to the problem. I would much rather simply solve for the number of half-lives and multiply that answer by the given length of the half-life.

3+=+10%280.5%5Ex%29
0.3+=+0.5%5Ex
log%28%280.3%29%29+=+x%2Alog%28%280.5%29%29
x+=+log%28%280.3%29%29%2Flog%28%280.5%29%29+=+1.7369656

t = 18.72*1.7369656 = 32.515995....

This method of solving the problem has a huge advantage if you are solving a large number of similar problems involving half-lives.

If the problems are with all different half-lives, then solving the problem by the first method shown involves a different first calculation for every problem; with the second method, the first calculation is always the same, and then you just need to multiply the result of that calculation by the half-life.

Someone who works this kind of problem often would undoubtedly work the problem this way.

For example, suppose we have an initial amount 23 and a final amount 5, with a half-life of 385 years. The calculations would be

(1) find the number of half-lives:
x+=+log%28%285%2F23%29%29%2Flog%28%280.5%29%29 = 2.201634....
(2) multiply by the half-life:
2.201634%2A385 = 847.63 years