\(M=m_o*\frac{1}{2}^\frac{t}{T}\\
M-mass\ aftter\ decay\\
m-initial\ mass\\
t-time\\
T-period(half-life)\\
25=200*\frac{1}{2}^\frac{3460}{T}|:200\\
\frac{1}{8}=(\frac{1}{2})^\frac{3460}{T}\\
(\frac{1}{2})^3=(\frac{1}{2})^\frac{3460}{T}\\
3=\frac{3460}{T}\\
T=1153\frac{1}{3}\)
or easiest way
200g -----> 100g -------> 50g -----> 25g
We can that there were three decays.
1 decay ----------- x years
3 decays---------- 3460
\(x=1153\frac{1}{3}\)
