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

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Steve S. | Tutoring in Precalculus, Trig, and Differential CalculusTutoring in Precalculus, Trig, and Diffe...
5.0 5.0 (3 lesson ratings) (3)
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
Jon-Erik R. | Effective Math and Science Tutoring (Elementary through College)Effective Math and Science Tutoring (Ele...
5.0 5.0 (77 lesson ratings) (77)
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:
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/t1/2; where t is the time (age) and t1/2 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/t1/2=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:
solving you get:
To find k, you simply take the natural log (ln) of both sides.
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.
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
Solving for t you get:
t=-ln(.46)/.00012  = 6471 years