Exponential problem: caffeine dosage

[jackscholar]

If the concentration of caffeine in a system at any given time is given by the equation
y(t)=De^-kt
where dy/dt=-kt is the clearence rate (re-arranged and integrated to form the above equation) and the concentration of caffeine in the system at t=0 is D, then calculate k if:
After one hour, 25% of the caffein has been cleared.

I know that y(60)= 3D/4 or three quarters of D
so how do I re-arrange to get k?

 

[Dick]

If the concentration of caffeine in a system at any given time is given by the equation
y(t)=De^-kt
where dy/dt=-kt is the clearence rate (re-arranged and integrated to form the above equation) and the concentration of caffeine in the system at t=0 is D, then calculate k if:
After one hour, 25% of the caffein has been cleared.

I know that y(60)= 3D/4 or three quarters of D
so how do I re-arrange to get k?

y(60)=De^(-k*60). That's equal to 3D/4. It shouldn't be too hard to solve for k if you use a log, is it?

 

[jackscholar]

OH! I see now. Just subtitute in 3/4D for y(60) then divide by D, take ln of both sides and divide by negative 60. Thank you!

 

[Ray Vickson]

If the concentration of caffeine in a system at any given time is given by the equation
y(t)=De^-kt
where dy/dt=-kt is the clearence rate (re-arranged and integrated to form the above equation) and the concentration of caffeine in the system at t=0 is D, then calculate k if:
After one hour, 25% of the caffein has been cleared.

I know that y(60)= 3D/4 or three quarters of D
so how do I re-arrange to get k?

Note: dy/dt is NOT equal to -kt; it is equal to -ky.