Question 763706
in 10 years, 25% of a radioactive substance decays.
 What is its half-life?
:
That means 75% remains
:
The radio active decay formula
A = Ao*2^(-t/h), where
A = amt remaining after t 
Ao = initial amt (t=0)
t = time of decay
h = half life of substance
:
Let initial amt = 1 then resulting amt = .75
1*2^(-10/h) = .75
ln(2^(-10/h)) = ln(.75)
{{{-10/h}}}*ln(2) = ln(.75)
{{{-10/h}}} = {{{ln(.75)/ln(2)}}}
{{{-10/h}}} = -.415
-.415h = -10
h = {{{(-10)/(-.415)}}}
h = +24 yrs is the half life