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Posted by **HM** on Monday, September 26, 2011 at 1:19pm.

How many years to the nearest year would be required for a given amount of this isotope to decay to 55% of that amount.

So I am not sure where to put the 1,300

1,300=2600*e^t

or A=x*e^1300

- MATH help -
**Steve**, Tuesday, September 27, 2011 at 11:06amAssuming a mass of Ao at time t=0, at time t=1300, m = 1/2

That is, A = Ao * 2^-(t/1300)

You see, that as t = 1300, we have 2^-1 = 1/2 Ao

So, now we need to convert 2^-n to e^-n

2 = e^(ln 2)

A = Ao * (e^(ln 2))^(-t/1300)

A = Ao * e^(-t/1300 * ln 2)

A = Ao * e^(-t/1875.5)

So, when A = 0.55 Ao, we have

.55 = e^(-t/1875.5)

ln(.55) = -t/1875.5

-0.5978 = -t/1875.5

t = 1121 years

Makes sense, since at t=1300 years, the amount will be reduced to 0.5

- MATH help -
**HM**, Wednesday, September 28, 2011 at 7:30amThank you Steve, thank you very much.

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