Suppose that after 3 hours 70% of the drug remains and also the drug decreases exponentially. How much time elapses until only 25% of the drug remains?

Question

Please help 

Answer 1

 

P(t) = percent of drug present at time t

 

P(t) = P(0)e^-kt where t is hours, P(0) = initial percent present = 100, and k has units of hours^-1

 

P(t) = 100e^-kt

 

To find k,

 

P(3) = 70 = 100e^-k(3)

 

k = ln(70/100) / (-3) = 0.11889 hours^-1

 

Finally, P(t) = 100e^-0.11889t

 

To find time T when P = 25,

 

25 = 100e^-0.11889T

 

T = ln(25/100) / (-0.11889) = 11.6603 hours

 

So by 11.6603 hours, the percentage has dropped to 25%.