Q:

A hiker in Africa discovers a skull that contains 32% of its original amount of C-14. find the age of the skull.

Accepted Solution

A:
Answer:9419.3 years.Step-by-step explanation:Let the initial amount of C-14 be 100 units. We have been given that a hiker in Africa discovers a skull that contains 32% of its original amount of C-14. We are asked to find the age of the skull.We will use half life formula to solve our given problem.[tex]A=a\cdot(\frac{1}{2})^{\frac{t}{h}}[/tex], where,A = Amount left after t years,a = Initial amount,t = time,h = Half life.We know that half-life of C-14 is 5730 years.32% of 100 units would be 32.[tex]32=100\cdot(\frac{1}{2})^{\frac{t}{5730}}[/tex][tex]\frac{32}{100}=\frac{100\cdot(\frac{1}{2})^{\frac{t}{5730}}}{100}[/tex][tex]0.32=(0.5)^{\frac{t}{5730}}[/tex]Now, we will take natural log of both sides.[tex]\text{ln}(0.32)=\text{ln}((0.5)^{\frac{t}{5730}})[/tex]Using log property [tex]\text{ln}(a^b)=b\cdot\text{ln}(a)[/tex], we will get:[tex]\text{ln}(0.32)=\frac{t}{5730}\cdot \text{ln}(0.5)[/tex][tex]\frac{\text{ln}(0.32)}{\cdot \text{ln}(0.5)}=\frac{t\cdot \text{ln}(0.5)}{5730\cdot \text{ln}(0.5)}[/tex][tex]1.64385618977=\frac{t}{5730}[/tex][tex]\frac{t}{5730}=1.64385618977[/tex][tex]\frac{t}{5730}*5730=1.64385618977*5730[/tex] [tex]t=9419.295967[/tex] [tex]t\approx 9419.3[/tex] Therefore, the age of skull is approximately 9419.3 years.