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Question

A scientist determined that the bones from a mastodon had lost 70.2?% of their carbon-14. How old were the bones at the time they were discovered?

Answer 1

Hi Logan,

 

There is some missing information here, and that is the half-life of carbon-14. I just looked it up, at it is 5,730 years.

 

The radioactive decay equation goes like

 

M(t) = M₀(1/2)^t/τ

 

where

 

M(t) is the mass of carbon-14 remaining as a function of time t,

M₀ is the initial mass of carbon-14,

τ = 5730 years, which is the half-life of carbon-14

 

We are told that the mastodon has lost 70.2% of its carbon-14. This means that it has 100% - 70.2% = 29.8% = 0.298 of it remaining. This means that M(t) / M₀ = 0.298.

 

Let's now solve the above equation for t.

 

We write

 

(1/2)^t/τ = M(t)/M₀

 

Then, we take the natural log of each side.

 

ln[(1/2)^t/τ] = ln[M(t)/M₀]

 

We know from the law of logarithms ln(a^b) = bln(a) that we can turn the power t/τ into a front coefficient:

 

(t/τ)ln(1/2) = ln[M(t)/M₀] 

 

Rearranging we get

 

t = τ ln=[M(t) / M₀] / ln(1/2)

 

Plugging in numbers, we get

 

t = (5730 years) ln(0.298)/ln(1/2)

 

So, t = 10,008 years