# SOLUTION: Will you please answer this question: An artifact was found and tested for its carbon-14 content. If 73% of the original carbon-14 was still present, what is its probable age (to

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 Question 470449: Will you please answer this question: An artifact was found and tested for its carbon-14 content. If 73% of the original carbon-14 was still present, what is its probable age (to the nearest 100 years)? Use that carbon-14 has a half-life of 5,730 years. Thank youAnswer by Theo(3464)   (Show Source): You can put this solution on YOUR website!73% of the original carbon-14 was still present. what is it's probably age to the nearest 100 years. carbon-14 has a half life of 5,730 years. here's a reference for half life formulas: http://math.ucsd.edu/~wgarner/math4c/textbook/chapter4/expgrowthdecay.htm the half life decay formula is: f = p * e^(-kt) where: f is the future value p is the present value e is the scientific base of the constant e whose value is 2.718281828. k is a constant whose value is positive which makes -k negative. t is the amount of time in years. in your problem, the half life is 5730 years. we use this half life to find the value of k. the formula is: 1/2 = 1 * e^(-kt) f is 1/2 p is 1 t is 5730 k is what we are trying to find. our formula becomes: .5 = e^(-5730k) to solve this problem, we use logarithms. in fact, this problem lends itself to natural logarithms, so we'll take the natural log of both sides of this equation to get: ln(.5) = ln(e^(-5730k) since ln(x^a) = a*ln(x), our formula becomes: ln(.5) = -5730k * ln(e) since ln(e) is equal to 1, our equation becomes: ln(.5) = -5730k we divide both sides of this equation by -5730 to get: ln(.5)/-5730 = k we solve for k to get: k = .000120968 we now have the value of k which we can use to solve the problem. we use our formula again, only this time we replace k with .000120968. f is the future value which is .73 times the original value which we assign as 1, which means the original value of 100% of the original value. our future value of .73 is 73% of the original value of 1. so our equation is: f = p * e^(-kt) this time we are solving for t which is the number of years. f = .73 p = 1 k = .000120968 the formula becomes: .73 = 1 * e^(-.000120968*t) this becomes: .73 = e^(-.00012068*t) since we want to solve for t which is in the exponent, we use natural logs again. we get: ln(.73) = ln(e^(-.00012068*t) which becomes: ln(.73) = -.00012068*t*ln(e) which becomes: ln(.73) = -.00012068*t because ln(e) is equal to 1. we divide both sides of this equation by -.00012068 to get: t = ln(.73) / -.00012068 to get: t = 2601.601245 years. we round this to the nearest hundred years to get: t = 2600 years. we can test our half life formula to see if it is accurate, by simply replacing .73 with .5 in that final equation to get: t = ln(.5) / -.00012068 to get: t = 5730. the equation is accurate and so we're good. the answer is that the carbon-14 is approximately 2600 years old. Note that in the reference they used N and N[0] and they used k. Their N is equivalent to my f. Their N[0] is equivalent to my p. Their k is equivalent to my -k because they state that, in the decay formula, the value of k is negative. My k is positive but it has a - sign in front of it, making the value negative.