# How to solve carbon 14 dating problem validating a database

How old is a skeleton that has lost 41% of its carbon-14?

Note: Do not round any numbers during your calculation. The radioactive element carbon-14 has a half-life of 5750 years.

section, i thought i would try here first to see if there was something obvious i missed...well, heres the question: Analysis on an animal bone fossil at an archeological site reveals that the bone has lost between 90%-95% of c-14.

Give an interval for the possible ages of the bone.

Note: Do not round any numbers during your calculation.

_____years old If you can please show all steps, thanks for any help!!! The amount of carbon-14 present decreases exponentialy with time.

The method of carbon dating makes use of the fact that all living organisms contain two isotopes of carbon, carbon-12, denoted 12C (a stable isotope), and carbon-14, denoted 14C (a radioactive isotope).

The ratio of the amount of 14C to the amount of 12C is essentially constant (approximately 1/10,000).

If we assume Carbon-14 decays continuously, then $$ C(t) = C_0e^, $$ where $C_0$ is the initial size of the sample. Since it takes 5,700 years for a sample to decay to half its size, we know $$ \frac C_0 = C_0e^, $$ which means $$ \frac = e^, $$ so the value of $C_0$ is irrelevant.ln(Ao/A) = k*t where Ao = starting level or concentration and A is the level or concentration after time t, and k is the rate constant. That means that after 5570 years the level or concentration has decreased from 1 to 1/2. alright, so there was this question in my math book, doesn't give many details, thats why i found it confusing...debating whether or not to post this question in chem.Students should be guided to recognize the use of the logarithm when the exponential function has the given base of $e$, as in this problem.Note that the purpose of this task is algebraic in nature -- closely related tasks exist which approach similar problems from numerical or graphical stances.

Now, take the logarithm of both sides to get $$ -0.693 = -5700k, $$ from which we can derive $$ k \approx 1.22 \cdot 10^.