Question 646647
The half-life of radioactive strontium-90 is approximately 29 years.
 In 1964, radioactive strontium-90 was released into the atmosphere during 
testing of nuclear weapons, and was absorbed into people's bones.
 How many years (since 1964) does it take until only 9 percent of the
 original amount absorbed remains?
:
The radio-active decay formula: A = Ao*2^(-t/h), where
A = resulting amt after t time
Ao = initial amt (t=0)
t = time of decay
h = half-life of substance
:
Let initial amt = 100 then resulting amt = 9
100*2^(-t/29) = 9
divide both sides by 100
2^(-t/29) = .09
using nat logs, the log equiv of exponents
{{{-t/29}}}*ln(2) = ln(.09)
{{{-t/29}}} = {{{ln.09/ln(2)}}}
{{{-t/29}}} = -3.47393
t = -29 * -3.47393
t = 100.744 yrs, plus 1964 = 2065 for bones to be reduced to 9%
:
:
:
That seemed like a long time, checked on a calc: 
enter 100*2^(-100.744/29) results: 9.000