The level of radioactivity on the site of a nuclear explosion is decaying exponentially. The level measured in 1970 was found to be 0.7 times the level measured in 1960. What is the half-life. (the amount of years)

2. Radioactivity in 1970 = $x$

Radioactivity in 1960 = $y$

But $x = 0,7y$

$0,7y = y(1 - \frac{r}{100} ) ^{10}$

$0,7 = (1 - \frac{r}{100} ) ^{10}$ [Divide by y]

$(1 - \frac{r}{100} ) = \sqrt[10]{0,7}$

$(1 - \frac{r}{100} ) = 0,9649$

$\frac{r}{100} = 0,0350$

$r = 3,503$

I think this is the rate at which it is decaying. I'll spend a little more time on it...

EDIT: Okay i got it! We now have the rate, so we want to know when $x = 0,5y$

$0,5y = y(1 - \frac{3,503}{100}) ^ {n}$

$0,5 = (1 - \frac{3,503}{100}) ^ {n}$

$n = log_{ 1 - \frac{3,503}{100} }{0,5}$

$n = 19,43 \ years$