A mummy discovered in Egypt has lost 46% of its carbon-14. Determine its age

## find using carbon

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# 2 Answers

A = P * (1/2)^(t/5730)

P = 100% = 1

A = 100% - 46% = 0.54

0.54 = 1*(1/2)^(t/5730)

Taking the log of both sides (base 10, but could do ln):

log(0.54) = log((1/2)^(t/5730))

Using power rule of logs:

log(0.54) = (t/5730)*log(0.5)

t = 5730 * log(0.54) / log(0.5)

t ≈ 5,094 years old

First, it is important to note that the half life of carbon-14 is 5730 years. This says that the amount of carbon is exponentially related to the length since decay began. To start, note that half life is represented by equation 1 below:

Equation 1:

A(t)=A*e

^{kt/t1/2}This says that the amount left is equal to the initial amount multiplied by e raise to the power of the k (the decay constant) and the number of half lives, or simply t/t

_{1/2; }where t is the time (age) and t_{1/2}is the half lift of the subject.If you are unfamiliar with the k constant for carbon-14, then it can be solved by plugging in a known value: the time of the half life, or t/t

_{1/2}=1. Since the time you plug in at this point is the half life of the substance, then A(t)=(1/2)*A.Then you have:

.5A=A*e

^{k}solving you get:

.5=e

^{k}To find k, you simply take the natural log (ln) of both sides.

ln(.5)=-.693

Now that you have k, it is a matter of substituting the amount of carbon-14 left at the age you would like to calculate.

So,

.46A=A*e

^{-.693*t/5730}Notice that the A's cancel out and you get:

.46=e

^{-.693t/5730 }or .46=e^{-.00012t}Take the natural log (ln) of both sides to eliminate e and you have

ln(.46)=-.00012t

Solving for t you get:

t=-ln(.46)/.00012 = 6471 years