t = 12040 years
N=NOe^-kt
N = amount of C-14 at time t
NO = initial amount of C-14
k= 0.0001
t = time (unknown)
We know that the amount found is 30% of the original. Therefore we can simply set N to 30 and NO to 100 because 30% of 100 is 30. Really you can give any amount, but setting it at 100 is the easiest, so long as N is 30% of NO.
so if we set the problem up again using all the given values and the new ones we assigned:
N= 30
NO = 100
30=100e^-0.0001(t)
Remember, you are solving for t, so you have to try and manipulate the formula to get t by itself on one side.
We can divide both sides by 100 and you are left with:
.30=e^-0.0001(t)
You can see that you have "e" which means we can use natural log to get rid of e:
e^y = x and ln(x) = y
.30=e^-0.0001(t)
using the natural log,
.30 = x
-0.0001(t) = y
Therefore:
ln (x) = y
ln (.30) = -0.0001(t)
plug in the value ln(.30) into a calculator and you get:
-1.20397 = -0.0001(t)
*NOTE: normally it would not work if you have a negative answer because you cannot have negative years, however, there is a negative on the other side of the equation which will cancel out so it works in this case.
now to solve for t, we simply divide both sides by -0.0001
Therefore
t = 12039.7
But you have to round to the nearest year so
t = 12040
**NOTE: the k value given (which is the rate at which C-14 decays each year) is very small. That means that you will have a large value for time because it takes so long for the C-14 to decay every year.
Hope that helped :)
