Question 867851
A certain radioactive substance decays according to the formula q(t)= q0e^-.0063t. where q0 is the initial amount of the substance and t is the time in days. Approximate the half-life of the substance.
:
let Qo = 1, then Q(t) = .5
A simple equation
{{{1*e^(-.0063t) = .5}}}
find the ln of both sides
{{{ln(e^(-.0063t)) = ln(.5)}}}
the log equiv of exponents
-.0063t*ln(e) = ln(.5)
the ln of e is one so we have
-.0063t = -.693147
t = {{{(-.693147)/(-.0063)}}}
t = +110 days is the half life of the substance