.5 = e^-k(10)
-.693 = -10 k
k = .0693
then
6/18 = e^-.0693 t
-1.0986 = - .0693 t
t = 15.9 years
The half life of a certain radioactive substance is 10 years. How long will it take for 18 gms of the substance to decay to 6 gms?
A. 6 ln 10 years
B. 10 ln 6 years
C. 10 (ln 3)/(ln 6) years
D. 18 ln 10 years
E. 10 (ln 3)/(ln 2) years
5 answers
I would solve
18(1/2)^(t/10) = 6
(1/2)^(t/10) = 1/3
from the answers, my clue would be to take ln of both sides
18(1/2)^(t/10) = 6
(1/2)^(t/10) = 1/3
from the answers, my clue would be to take ln of both sides
Thanks I think it would be E.10 (ln3)/(ln2) years
Right?
Right?
10 ln 3/ln 2 = 15.9 yes
P = P(0)e^(kt)
(1/2) = e^(10k)
ln(1/2) = 10k
k = (1/10)ln(1/2)
----------------
6 = 18e^((1/10)ln(1/2)t)
(1/3) = e^((1/10)ln(1/2)t)
t = 10[ln(1/3)/ln(1/2)]
t = 10[-ln(3)/-ln(2)]
t = 10[ln(3)/ln(2)]
(1/2) = e^(10k)
ln(1/2) = 10k
k = (1/10)ln(1/2)
----------------
6 = 18e^((1/10)ln(1/2)t)
(1/3) = e^((1/10)ln(1/2)t)
t = 10[ln(1/3)/ln(1/2)]
t = 10[-ln(3)/-ln(2)]
t = 10[ln(3)/ln(2)]