Answer:
This is an exponential decay problem. These problems are of the form:
y = c * ekt
y = amount of substance left
c = original amount of substance (Here, set this to 100%, or 1.)
e = exponential constant (~2.718)
k = rate of decay constant (We need to figure this out.)
t = time, in years (When t = 0, y = 1 (all of the carbon-14). When t = 5730, y = 0.5 (one half-life has passed). When t = the answer we're trying to find, y = 0.98 (98% of the carbon-14).)
First, we must find the value of k, the rate of decay constant. We know that after 5730 years (t = 5730), one-half of the carbon-14 will remain (y = 0.5).
y = c * ekt
0.5 = 1 * ek * 5730
0.5 = e5730k
To get rid of the e, take the natural logarithm (ln) of both sides:
ln(0.5) = ln(e5730k)
ln(0.5) = 5730k
ln(0.5)/5730 = k
ln(0.5) is a negative number, so our rate of decay constant will be negative. This is a little "sanity check", because radioactive decay means the amount of substance goes down over time. If you get a positive value of k, then you made a mistake somewhere.
Now that we have the rate of decay constant, we can find the value of t (in years) that will yield 98% of the carbon-14 remaining.
y = c * ekt
0.98 = 1 * e[ln(0.5)/5730]t
0.98 = e[ln(0.5)/5730]t
To get rid of the e, take the natural logarithm (ln) of both sides:
ln(0.98) = ln(e[ln(0.5)/5730]t)
ln(0.98) = [ln(0.5)/5730]t
ln(0.98) = ln(0.5)t / 5730
Solve for t:
5730 * ln(0.98) = ln(0.5)t
5730 * ln(0.98) / ln(0.5) = t
Put that into your calculator, to get t ~ 167 years. Both ln(0.98) and ln(0.5) are negative, so the negatives will cancel out to yield a positive number (another "sanity check").
Explanation: