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97% steady state after 5.5 half lives
t(1/2)=(.7xVd)/CL VERY IMPORTANT FORMULA
page 195 in F.A 2006
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ok...I understand how you got the 97% but what did u use the formula for?
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i have a hard time to get this topic, would you guys explain?
thanks,
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50% of half life is achieved in 1 half life.
75% is acheived in 2 half lives.
87.5% is acheived in 3 half lives. (to simplify some calculations follow that 90% is achived in 3.3 half lives).
95% is achieved in 4-5 half lives.
100% in more than 7 half lives.
Clinically steady state conc. is reached in 4-5 half lives. So for a drug with half live of 4 hrs the stady state will be achieved in 4*5= 20 hrs.
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drrkamra, you are right, but just to be exact since sometime NBME ask those things,
94% is achieved in 4hl,
97% is achieved in 5hl,
99% is ..... in 6hl.
IT takes>7 half life to reach real ss, but clinical ss is considered at 4-5 hl (t1/2).
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Thanks for you replies !!!'
ok.....so let me get this straight.... no matter what the half life is the steady state will always be 4-5hrs times the half life?
In the question that I had asked removal of the kidney did not make any difference? Because when you removed one kidney the half life is 8 hours? So how do you still end up with 20 if 8x5 = 40?
any thoughts