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March 1, 2015

March 1, 2015

Posted by **james** on Tuesday, January 22, 2013 at 10:15pm.

- precal -
**Reiny**, Tuesday, January 22, 2013 at 10:20pmFor half-life question, a good equation to use is

amount = starting value (1/2)^(t/k) where k is the half-life

so :

.35 = 1(1/2)^(t/2000) , t in years

.35 = (.5)^(t/2000

log .35 = log ( (.5)^(t/2000)

log .35 = (t/2000) log .5

t/2000 = log .35/log .5

t = 2000( log .35/log .5) = 3029.14

it would take appr 3029 years

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