The level of radioactivity on the site of a nuclear explosion is decaying exponentially. The level measured in 1980 was found to be 0.8 times the level measured in 1970. What is the half-life.
Let L(0) denote the initial level of radiation and t the passed time. Then the level after t years is calculated by:
$\displaystyle L(t)=L(0)\cdot e^{k\cdot t}$ with k a constant typical for the decaying process.
From L(0) = 1 and L(10) = 0.8 you can derive k:
$\displaystyle 0.8 = 1\cdot e^{k\cdot 10}~\implies~ k=\frac1{10} \cdot \ln(0.8)~\approx~-0.02231...$
Now calculate the time which is necessary to reach only the half of the radioactivity:
$\displaystyle 0.5 = e^{\frac1{10} \cdot \ln(0.8) \cdot t}~\implies~ t\approx 31.06\ yrs$
Btw: That is a very optimistical result
$\displaystyle A(t)=A_0\left( \frac{1}{2}\right)^{\frac{t}{h}}$
let 1970 be time t=0
Then
$\displaystyle A(10)=.8A_0=A_0\left( \frac{1}{2}\right)^{\frac{10}{h}}$
$\displaystyle .8=\left( \frac{1}{2}\right)^{\frac{10}{h}}$
$\displaystyle \ln(.8)=\frac{10}{h}\ln\left( \frac{1}{2}\right)$
$\displaystyle h=\frac{10 \ln \left( \frac{1}{2}\right)}{\ln(.8)} \approx 31.06$
I hope this helps.