Asked by Anonymous
                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
            
        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
Answers
                    Answered by
            Damon
            
    .5 = e^-k(10)
-.693 = -10 k
k = .0693
then
6/18 = e^-.0693 t
-1.0986 = - .0693 t
t = 15.9 years
    
-.693 = -10 k
k = .0693
then
6/18 = e^-.0693 t
-1.0986 = - .0693 t
t = 15.9 years
                    Answered by
            Reiny
            
    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
                    Answered by
            Anonymous
            
    Thanks I think it would be E.10 (ln3)/(ln2) years
Right?
    
Right?
                    Answered by
            Damon
            
    10 ln 3/ln 2 = 15.9 yes
    
                    Answered by
            Austin
            
    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)]
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