# PharmPK Discussion - Defining 0.693 correction factor for converting Ke to t1/2

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• On 3 Dec 2001 at 14:26:05, "Gendron, Richard T. (GS)" (RTGendron.at.mar.med.navy.mil) sent the message
`Please post to pkin website:I am occasionally asked to explain why 0.693 is used to convert theelimination rate constant to half-life. I have been offering the followingexplanation. Please tell me if I am correct in making the followingassumption:The factor 0.693 "unlogs" the elimination rate constant to yield half-lifeby the following relationship: 1/e^0.693 = 50%.If I am incorrect, please someone set me straight.Thanks.R. Timothy GendronPharmacy DepartmentNaval Medical Center27 Effingham StreetPortsmouth, VA 23708Phone (757) 953-0252Fax (757) 953-3328`
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• On 4 Dec 2001 at 10:55:52, David Bourne (david.aaa.boomer.org) sent the message
`[A few replies - I'm rather partial to the second but then..:-) - db]=46rom: "JVIS (Jennifer Visich)" Date: Mon, 3 Dec 2001 13:51:22 -0800To: david.-at-.boomer.orgSubject: RE: PharmPK Defining 0.693 correction factor for convertingKe to	 t1/2The following message was posted to: PharmPKRichard,converting by 0.693 is a result of the following	the first order equation:	C(t)=3D C(0)e^-kt	rearranged you get: 	C(t)/C(0) =3D e^-kt	when C(t) is one half of C(0)you get:					0.5 =3D e^-kt	take the natural log of both sides to get:					-.693 =3D -kt	rearranged					t1/2=3D.693/kHope that helps,Jenn Visich, Ph.D.Sr ScientistZymoGenetics, Inc.1201 Eastlake Ave ESeattle, WA 98102(206) 442-6728jvis.-a-.zgi.com---=46rom: Stephen Day Date: Mon, 3 Dec 2001 17:29:08 -0500 (EST)To: david.-a-.boomer.orgSubject: Re: PharmPK Defining 0.693 correction factor for converting Ke to t=1/2The following message was posted to: PharmPKRichard,David Bourne has a nice explanation here:http://www.boomer.org/c/p1/All you need to know is in section 4 "One CompartmentI.V. Bolus"0.693 is the natural logarithm of 2, almost (i.e.,ln(2)=3D 0.693147....)Steve---=46rom: "Thomas A Torda" Date: Tue, 4 Dec 2001 10:19:09 +1100To: david.-a-.boomer.orgSubject: Re: PharmPK Defining 0.693 correction factor for converting Ke to t=1/2Status: RThe following message was posted to: PharmPKPerhaps, simpler migt be:1    Assuming a linear fall in log(concentration) v time, otherwise 1/2lives do not exist:Using the Naperian (natural) logs, the equation for the ln(conc), y, is y ==3DC(0 time) + b.time (b will be negative and is the rate constant ofelimination or whatever, the slope of the concentration line.)2    1/2 life =3D time for halving of concentration, that is, time for conc(=2)=3D conc(1)/2 or ln(conc(1)) - ln(conc(2)) =3D ln 2 =3D 0.6930.693 =3D (Co + b.time1) - (Co + b.time2)           =3D b(time1 - time2)3      1/2 life or 1/2 time =3D  0.693/bTom Torda---=46rom: "samia ezzine" Date: Mon, 03 Dec 2001 23:03:04To: david.aaa.boomer.orgSubject: Re: PharmPK Defining 0.693 correction factor for converting Ke to t=1/2Status: RThe following message was posted to: PharmPKHi Timothy,As Cp =3D Coe-ketAt t1/2, Cp =3D C0/2 and t=3D t1/2  yields to C0/2 =3D C0 e-ket1/2by simplifying  e-ket1/2 =3D 1/2 ,  Ket1/2 =3D Ln 2 sot1/2 =3D Ln2/ke and Ln 2=3D 0.693 so t1/2 =3D 0.693/keI hope that this can help you.Samia Ezzineph.D student=46acult=E9 de pharmacieUniversit=E9 de Montr=E9alQc, Canada---=46rom: "Bruce CHARLES" Date: Tue, 4 Dec 2001 10:23:56 10To: david.at.boomer.orgSubject: Re: PharmPK Defining 0.693 correction factor for converting Ke to t=1/2Status: RThe following message was posted to: PharmPKSend reply to:  	PharmPK.at.boomer.orgDate sent:      	Mon, 3 Dec 2001 14:26:05 -0600=46rom:           	"Gendron, Richard T. (GS)" (by way of David Bourne)To:             	Multiple recipients of PharmPK - Sent bySubject:        	PharmPK Defining 0.693 correction factor forconverting Ke to t1/2Going back a step further, define half-life (t1/2) as time to go fromconcentration 2C (at time, t2) to C (at time, t)k (=3Dslope)   =3D Ln (2C) - Ln(C) / (t2-t)                     =3D Ln (2C/C) / t1/2                     =3D Ln 2 / t1/2               k    =3D 0.693 / t1/2q.e.d.Cheers,BC---From: Michael.D.Karol.aaa.abbott.comDate: Tue, 4 Dec 2001 10:08:09 -0600MIME-Version: 1.0Richard:In pharmacokinetics, we refer to half-life, the time for half of theelimination process to be completed.  The value of 0.693 is ln(2), reciproca=lof 1/2.Thus, we have K =3Dln(2)/ "half life"   =3D 0.693/ "half life".In physics or engineering, the term "time constant" is used.  This is the ti=mefor 1/e to be completed (where e represents the base of natural log).To convert from a time constant to rate constant the following would apply:K =3D ln(e)/ "time constant".   Note that ln(e) =3D 1.Other constant times could be used to describe first order processes.  Forexample, you could use the time for one third of the elimination to becompleted.  This might be called a third life (as opposed to a half life).  =Inthis case the equation would be:K =3D ln(3)/ "third life"Hope this helps.MK---=46rom: Glenn Whelan Date: Tue, 4 Dec 2001 11:27:05 -0500To: david.-a-.boomer.orgSubject: RE: PharmPK Defining 0.693 correction factor for convertingKe to	 t1/2The following message was posted to: PharmPKDr. Gendron:0.693 comes from the natural log of 2 thus to find time, t, for half of aconcentration, Xo:0.5Xo =3D Xo e-kt1/2(0.5Xo/Xo)=3De-kt1/2    Xo cancels outLn(0.5)=3D -kt1/20.693=3D -kt1/20.693/-k =3D t1/2Glenn Whelan, PharmD, Research FellowNemours Children's Clinic807 Children's WayJacksonville, FL 32250Phone 904.390.3406=46ax 904.390.3425email gwhelan.-a-.nemours.org`
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• On 5 Dec 2001 at 13:48:06, David Bourne (david.-at-.boomer.org) sent the message
`[Three more replies - db]From: David Jaworowicz Date: Tue, 04 Dec 2001 12:23:57 -0500To: david.-a-.boomer.orgSubject: Re: PharmPK Defining 0.693 correction factor for converting Ke to t1/2A more straightforward way of explaining where 0.693 comes from and why it'sused:For a drug exhibiting monoexponential disposition: Cp=Co*e^(-k*t)where Co is initial drug concentration, Cp is drug at time t, and k is theelimination rate constant.Rewritten:  Cp/Co=e^(-k*t)At t=t1/2 (i.e half-life):  Cp/Co=0.5  and  thus 0.5=e^(-k*t1/2)Solving for t1/2:  ln 0.5 = -0.693 = -k * t1/2and thus t1/2= 0.693/kSincerely,David Jaworowicz---From: "Daniel Sitar" Organization: University of ManitobaTo: david.at.boomer.orgDate: Tue, 4 Dec 2001 11:44:00 -0600Subject: Re: PharmPK Re: Defining 0.693 correction factor forconverting Ke to t1/2Reply-to: sitar.aaa.Ms.UManitoba.CAX-Confirm-Reading-To: sitar.at.ms.UManitoba.CAX-pmrqc: 1Priority: normal	Many students don't follow the math presented in thesesolutions.  It is better showgetting rid of the "-" sign by taking the reciprocal of e^-kt.  Then1/e^kt = 0.5e^ kt = 2Ln of both sides   yields    kt = 0.693Dan Sitar, Pharmacology & TherapeuticsUniversity of Manitoba---From: "Ronald A. Herman" Date: Tue, 4 Dec 2001 12:54:31 -0600To: david.-a-.boomer.orgSubject: RE: PharmPK Re: Defining 0.693 correction factor forconverting Ke to t1/2The following message was posted to: PharmPKAs is accurately pointed out, Ln 2 = 0.693147...... so let's all stop using0.693 and start using Ln 2.  Why add rounding error? Ln 2 is on mycalculator and only takes me two keystrokes whereas, 0.693 is 5 keystrokes!Ron HermanRonald A. Herman, Ph.D.   Asst. Professor (Clinical), Director   Iowa Drug Information Network   University of Iowa College of Pharmacy`
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