How old is a bone that has 12.5% of the original amount radioactive carbon 14 remaining?

Nuclear decay formula is N (t) = N₀*2^ - (t/T), where N (t) is the amount of nuclear material in some moment t, N₀ is the original amount of nuclear material, t is time and T is the half life of the material, in this case carbon 14. In our case N (t) = 12.5% of N₀ or N (t) = 0.125*N₀, T=5730 years and we need to solve for t:

0.125*N₀=N₀*2^ - (t/T), N₀ cancels out and we get:

0.125=2^ - (t/T),

ln (0.125) = ln (2^ - (t/T))

ln (0.125) = - (t/T) * ln (2), we divide by ln (2),

ln (0.125) / ln (2) = - t/T, multiply by T,

{ln (0.125) / ln (2) }*T=-t, divide by (-1) and plug in T=5730 years,

{ln (0.125) / [-ln (2) ]}*5730=t

t=3*5730=17190 years.

The bone is t = 17190 years old.