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September 30, 2014

September 30, 2014

Posted by **Hannah** on Saturday, November 20, 2010 at 4:54pm.

Would I use the formula N=No e^kt?

- Math -
**drwls**, Saturday, November 20, 2010 at 5:09pmYou could use that formula, if you know how to relate k to half life. In the formula you wrote, there would have to be a negative exponent (assuming k is positive).

Another way that might be easier to remember is to solve:

0.45 = (0.5)^(t/T)

where T is the half life.

Take logs of both sides (to any base)

t/T = log(0.45)/log(0.5) = 1.15

T = 5600/1.15 = 4870 years

- Math -
**Reiny**, Saturday, November 20, 2010 at 5:20pmyes, you could

You will need

.5 = 1 e^(5600k)

5600k = ln .5

k = ln .5 / 5600

so .45 = e^((ln.5/5600)t

(ln.5/5600)t = ln.45

t = 5600(ln.45)/ln.5

= 6451.2

or the other formula

N = (.5)^(t/5600)

then .45 = .5^(t/5600)

t/5600 = ln.45/ln.5

t = (5600ln.45/ln.5) = same calculation

- Math -
**Hannah**, Sunday, November 21, 2010 at 11:44amThank You

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