story problem? Calculus is not elementary school.
Q(t)=Qo*ekt
25=100*e12k
ln of both sides...
ln(1/4)=12k
-1.38629436=k*12
solve for k
then, for half life...
.5=ek*t
solve for t (take ln of both sides)
thalf=ln(.5)/k
A radioactive element decays exponentially according to the function Q(t) = Q0ekt
.If 100 mg of the element decays to 25 mg in 12 days, find the half-life of the element.
3 answers
Did not need the attitude bob pursley
Plug in the values given from the problem into Q(t) = Q0e^kt (Q(12) = 25, Q0 = 100, t = 12) and then solve for k
25 = 100e^(12k)
ln(1/4) = lne^(12k)
ln(1/4) = 12k
ln(1/4)/12 = k
Next, set Q(t) = 50 (since 50 is half of 100) and plug in the previously found value of k to solve for t
50 = 100e^(ln(1/4)/12)t
1/2 = e^(ln(1/4)/12)t
ln(1/2) = (ln(1/4)/12)t
12ln(1/2) = ln(1/4)t
12ln(1/2)/ln(1/4) = t
t = 6
The half-life of the element is 6
25 = 100e^(12k)
ln(1/4) = lne^(12k)
ln(1/4) = 12k
ln(1/4)/12 = k
Next, set Q(t) = 50 (since 50 is half of 100) and plug in the previously found value of k to solve for t
50 = 100e^(ln(1/4)/12)t
1/2 = e^(ln(1/4)/12)t
ln(1/2) = (ln(1/4)/12)t
12ln(1/2) = ln(1/4)t
12ln(1/2)/ln(1/4) = t
t = 6
The half-life of the element is 6
2 answers