Posted by Autumn on October 09, 2002 at 15:51:10:
In Reply to: Re: Having a hard time with logorhythms and e function... posted by Subhotosh Khan on October 09, 2002 at 15:08:22:
: : Hi there! I have a problem I'm trying to solve, and I'm having a hard time with it.
: : The problem is: The concentration y of a drug in a person's system decreases according to the relationship: y = 2e^-.125t
: : where y is in appropriate units, and t is in hours. Find the amount of time that it will take for the concentration to be half of its original value.
: : How would you set this problem up? To me, it seems there are two variables, y and t. The book gives the answer 5.5 hours, and if I set it up as:
: : 1=2e^-.125t
: : 1/2=e^-.125t
: : in.5=-.125t
: : in.5/-.125=t
: : then I get the answer 5.5
: : I don't understand why this works to use 1 for y...I thought to find 1/2 the concentration, I would need to use .5y and if I assume y is=1, then the problem would look like: .5(1)=2e^-.125t
: : .5=2e^-.125t
: : and so on, which gives me a wrong answer. I would like to understand how this works. Sorry if this is confusing...if it is, it's because I'm confused! Thanks!
: Say your original concentration at time t_1 was y_1
: y_1 = 2e^[(-t_1)/8].............(1)
: Now at time t_2 the concentration reduces to y_2
: y_2 = 2e^[(-t_2)/8] ..............(2)
: You are given (y_2)/(y_1) = 1/2
: You need to find t_2 - t_1 = ????
: Divide eqn(2) by eqn(1)
: (y_2)/(y_1) = 2e^[(-t_2)/8]/2e^[(-t_1)/8]
: = e^[(-t_2 + t_1)/8]
: 1/2 = e^[(-t_2 + t_1)/8]
: ln(1/2) = (-t_2 + t_1)/8
: Finish it from here...
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