Question

The bones of a prehistoric man found in the desert of new Mexico contain approximately 5% of the original amount of carbon 14. If the half-life of carbon 14 is 5600 years, approximately how long ago did the man die?

Answer 1

`~~8678.5` years ago

Explanation:

Quick note: The half life of C-14 is more commonly 5730 years, the value I will be using.

To find the age of an object with a radioactive element still present, we use this formula:
`t=(t_(1/2)ln(N_t/N_0))/-ln2`, where `t` is the age of the object, `t_(1/2)` is the half life of the element, `N_0` is the initial quantity of the element (usually 100), `N_t` is the remaining quantity of the element after time, and `ln` is the natural logarithm (base of `e`).
http://mathcentral.uregina.ca/beyond/articles/ExpDecay/Carbon14.html

As you can see, we have every value except for `t`. Plug our known variables into the equation, and we get
`t=(5730ln(35/100))/-0.693`
`t=((4011log)/2)/-0.693`
`t~~8678.5`

Therefore the bones of the prehistoric man are roughly 8678.5 years old.