SOLUTION: A piece of charcoal is found to contain 30% of the carbon 14 that it originally had. When did the tree die from which the charcoal came? Use 5600 years as the half-life of carbon 1

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: A piece of charcoal is found to contain 30% of the carbon 14 that it originally had. When did the tree die from which the charcoal came? Use 5600 years as the half-life of carbon 1      Log On


   



Question 250880: A piece of charcoal is found to contain 30% of the carbon 14 that it originally had. When did the tree die from which the charcoal came? Use 5600 years as the half-life of carbon 14.
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
formula is:

F = P*e^(kt) where:

F = future amount
P = present amount
k = constant of proportion.
t = time

half life of carbon is given by the equation:

.5 = e^(5600k) because:

half life of carbon is 5600 years.

solve for k and then you can solve the problem.

take log of both sides of equation of .5 = e^(5600k) to get:

log(.5) = log(e^(5600k))

this becomes:

log(.5) = 5600k * log(e)

divide both sides of this equation by 5600*log(e) to get:

k = log(.5)/(5600*log(e))

solve for k to get:

k = -.301029996 / (5600*(.434294482) which becomes:

k = -.000123776

substitute in your original equation to confirm this value for k is good.

.5 = e^(5600*k) becomes:

.5 = e^(5600*-.000123776) which becomes:

.5 = .5 confirming the value for k is good.

now that you have the value of k, you can solve the equation.

if the tree has only 30% of the carbon left, then your equation of:

F = P * e^(kt) becomes:

.3 = 1*e^(k*t) which becomes:

.3 = e^(-.000123776*t)

take the log of both sides of this equation to get:

log(.3) = log(e^(-.000123776*t)) which becomes:

log(.3) = -.000123776*t*log(e)

divide both sides of this equation by -.000123776*log(e) to get:

t = log(.3) / (-.000123776*log(e))

solve for t to get:

t = 9727.007327 years

to confirm this is correct, substitute in the original equation to get:

.3 = e^(-.000123776 * 9727.007327) which becomes:

.3 = .3

the key is knowing the formula f = p*e^(kt) which I hope they gave you.

without the use of the scientific constant of e, you would have had to at least know that the rate of growth or decay is exponential.

solving this without the use of the formula f = p*e^(kt) is possible using exponential growth formula of:

f = p*(1+x)^t

this is a compounding formula where x is the rate of growth or decay.

using this formula, you would get:

f = p * (1+x)^t becomes:

.5 = 1 * (1+x)^5600 which becomes:

.5 = (1+x)^5600

take the 5600th root of both sides of this equation to get:

(1+x) = .5^(1/5600) which becomes:

(1+x) = .999876231

now that you know what (1+x) is equal to, you can solve the equation.

equation of:

f = p*(1+x)^t becomes:

.3 = 1*(.999876231)^t which becomes;

.3 = (.999876231)^t.

take log of both sides of this equation to get:

log(.3) = log(.999876231^t) which becomes:

log(.3) = t*log(.999876231)

divide both sides of this equation by log(.999876231) to get:

t = log(.3) / log(.999876231)

solve for t to get:

t = 9727.007361 years.

that's pretty close to t = 9727.007327 years which is what we got using the exponential formula of f = p*e^(kt)

bottom line is:

your answer is 9727.007327 years assuming exponential rate of growth or decay.