calculus question

Question

The amount of​ carbon-14 present in animal bones after t years is given by

​P(t)equals=Upper P 0 e Superscript negative 0.00012 tP0e−0.00012t.

A bone has lost 24​% of its​ carbon-14. How old is the​ bone?

Answer 1

Thank you for your question Jennifer!  

In the framework of the question

P(t)=P0​ * e^(−0.00012t)

You can actually make the additional substitution that the new value will be 0.76*P0 because it’s losing 24% of it’s value (1.00 - 0.24 = 0.76)

That makes the equation

0.76*P0 = P0 * e^(-0.00012t)

0.76 = e^(-0.00012t)

To solve this expression then I would take the natural log of both sides in order to get the “e” to cancel

ln(0.76) = -0.00012t

t = ln(0.76) / -0.00012

t = 2286.97

rounding up to the nearest year that means that the bone is 2287 years old