Really confusing math....?

Question

Estimate the percentage of polonium (^210 Po) that remains in a sample after 2 years. Round to the nearest hundredth of a percent. (note...half life of 138 days)

also....

The Rhind papyrus contains most of what we know today of ancient Egyptian mathematics. A chemical analysis of a sample from the papyrus has shown that it contains approx. 75% of its original carbo-14. Estimate the age of the Rhind papyrus.

Answer 1

2 years is 730 days, so, assuming the polonium was 100% pure,
p = (0.5)^(730/138) = 0.02556, or 2.56%

the half life of c14 is 5730 years (http://en.wikipedia.org/wiki/Carbon_14), so we solve
0.75 = (0.5)^(t/5730)
log(0.75) = (t/5730) log(0.5)
t = 5730 log(0.75) / log(0.5)
t = (and the age should be less than the half life, since we have more than half of it left)
t = 2378.16 yr.

Answer 2

the half life is the time it takes for 1/2 of something to decay

in 138 days, 1/2 of the material will be left (1/2 decayed)
in another 138 days, 1/4 of the material will be left (another 1/2 decayed
in another 138 days, 1/8 of the material will be left (another 1/2 decayed

Notice that the amount of remaining material is given by
(1/2)^x, where x is the number of "half-life" periods

In the above examples, we have
(1/2)^1 = 1/2
(1/2)^2 = 1/4
(1/2)^3 = 1/8

For your case, assuming 365 days in a year -- (2 * 365) / 138 = 5.2898550724637681159420289855072. . .

Plug (1/2)^5.2898550724637681159420289855072 into a calculator to get
0.025562007060829363258337435343405. . .

I don't know the half life of carbon-14, so I can't do the second part, but it would be done similarly. . .