Not sure the precise concept of "normal observation", but I assume that is observed by "eyes".
Eye observation is basically macroscopic, but when you use a mark, which can be regarded as a point of mass, then it goes to microscopic.
Mark is a reference point which you can compare the relative position change, but with your eyes, first you cannot notice microscopic changes, second the eyes cannot precisely set a stable reference point.
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.
