Exponential and logarithmic functions

Question

Carbon 14 also known as radiocarbon, is a radioactive form of carbon that is found in all living plants and animals. the amount of radio carbon present, y, in animal bones after t years is given by y=y₀e^-.00012t where y₀ is the amount present in living animals.

A)find the half life ( round to nearest tenth)

B) a bone contains 14% of its carbon -14 estimate the age of the bone (round to nearest tenth)

Answer 1

In this type of equation, y₀ represents the initial amount of radiocarbon present. You can tell by plugging in t = 0:

 

y = y₀e^-0.00012(0)

y = y₀e⁰

y = y₀(1)

y = y₀

 

So, initially (at t = 0), the amount of radiocarbon present is y₀. We'd like to know the half-life, which is the time t after which only half of the initial amount is left. Plug in y = 1/2y₀ and solve for t:

 

1/2y₀ = y₀e^-0.00012t

1/2 = e^-0.00012t

ln(1/2) = -0.00012t

-0.6931 = -0.00012t

5776.23 = t

 

That means that the half-life of Carbon-14 is approximately 5,776.2 years.

 

 

 

For Part (b), we'd like to know how long it will be until 14% of the initial amount of radiocarbon in a bone remains. That means we want to know t when y = 0.14y₀. Plug that in for y and solve for t:

 

0.14y₀ = y₀e^-0.00012t

0.14 = e^-0.00012t

ln(0.14) = -0.00012t

-1.966 = -0.00012t

16384.27 = t

 

So, the age of the bone is approximately 16,384.3 years.