Question

The amount of​ carbon-14 present in animal bones after t years is given by ​P(t)equalsUpper P 0 e Superscript negative 0.00012 t. A bone has lost 19​% of its​ carbon-14. How old is the​ bone?

Answer 1

Answer:

age of bone is 1756 years

Step-by-step explanation:

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

$P(t)= P_0e^{-0.00012t}$

P(t) is the carbon present

19% has lost. so carbon present is 100-19 = 81% present

out of 100 81 is presents

so P0 is 100

P(t) is 81

$81= 100e^{-0.00012t}$

divide by 100 on both sides

$0.81= e^{-0.00012t}$

take ln on both sides

$ln(0.081)=-0.00012tln(e)$

$ln(0.081)=-0.00012t$

divide both sides by  -0.00012

t=1756.0085

age of bone is 1756 years