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March 30, 2015

March 30, 2015

Posted by **Peter** on Friday, February 17, 2012 at 1:03am.

- Algebra 2 Decay -
**DrBob222**, Friday, February 17, 2012 at 1:32amThis may not be the math process you want. From a chemists standpoint, I would do the following.

ln(No/N) = kt. We must determine k.

So at zero time we start with 20 g. At the end of 1 day we have 20g-5% = 19 g. Plug that into the first order equation of ln(20/19) = k(1 day)

k = 0.0513

Then ln(20/10) = 0.0513(t)

Solve for t. If I didn't goof that is

13.51 days which I would round to 13.5 days.

- Algebra 2 Decay -
**Reiny**, Friday, February 17, 2012 at 7:56amUsing the "continuous" decay formula, I got

.5 = 1(e^(-.05t) )

ln .5 - -.05t lne, but ln e = 1

t = -.05/ln .5 = .072134.. years

= 26.3 days

- Algebra 2 Decay -
**DrBob222**, Friday, February 17, 2012 at 11:37amI looked up the half life of P-32. It is 14.28 days. I suspect 5% decay in the problem is just a close number.

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