If [tex]S(t)[/tex] is the amount of strontium-90 present in the area in year [tex]t[/tex], and it decays at a rate of 2.5% per year, then
[tex]S(t+1)=(1-0.025)S(t)=0.975S(t)[/tex]
Let [tex]S(0)=s[/tex] be the starting amount immediately after the nuclear reactor explodes. Then
[tex]S(t+1)=0.975S(t)=0.975^2S(t-1)=0.975^3S(t-2)=\cdots=0.975^{t+1}S(0)[/tex]
or simply
[tex]S(t)=0.975^ts[/tex]
So that after 50 years, the amount of strontium-90 that remains is approximately
[tex]S(50)=0.975^{50}s\approx0.282s[/tex]
or about 28% of the original amount.
We can confirm this another way; recall the exponential decay formula,
[tex]S(t)=se^{kt}[/tex]
where [tex]t[/tex] is measured in years. We're told that 2.5% of the starting amount [tex]s[/tex] decays after 1 year, so that
[tex]0.975s=se^k\implies k=\ln0.975[/tex]
Then after 50 years, we have
[tex]S(50)=se^{50k}\approx0.282s[/tex]
