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. 

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