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# math help explain?

The half-life of 14C is 5730 years. A chemist measures the amount of 14C in a bone tissue sample to be about 1/16th of the amount that was present when it died. About how many years ago did the animal die?

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- AlanLv 71 month ago
A = A_0*(1/2)^(t/5730)

is A_0 is the original amount A/A_0 = (1/2)^(t/5730) and you are saying A/A_0 = 1/16 1/16 = (1/2)^(t/5730) take ln of both sides ln (1/16) = (t/5730)ln(1/2) (t/5730) = ln(1/16)/ln(1/2) t = ln(1/16)*5730/ln(1/2) = 22920 years

or alternatively

1/16 = (1/2)^(t/5730)

take log base 2 of both sides

log_b2(1/16) = (t/5730) log_b2(1/2)

-4 = (t/5730) * -1

-4/-1 = t/5730

t/5730 =4

t = 5730*4 = 22920 years

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