- On 3 Dec 2001 at 14:26:05, "Gendron, Richard T. (GS)" (RTGendron.at.mar.med.navy.mil) sent the message

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Please post to pkin website:

I am occasionally asked to explain why 0.693 is used to convert the

elimination rate constant to half-life. I have been offering the following

explanation. Please tell me if I am correct in making the following

assumption:

The factor 0.693 "unlogs" the elimination rate constant to yield half-life

by the following relationship: 1/e^0.693 = 50%.

If I am incorrect, please someone set me straight.

Thanks.

R. Timothy Gendron

Pharmacy Department

Naval Medical Center

27 Effingham Street

Portsmouth, VA 23708

Phone (757) 953-0252

Fax (757) 953-3328 - On 4 Dec 2001 at 10:55:52, David Bourne (david.aaa.boomer.org) sent the message

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[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 -0800

To: david.-at-.boomer.org

Subject: RE: PharmPK Defining 0.693 correction factor for converting

Ke to t1/2

The following message was posted to: PharmPK

Richard,

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/k

Hope that helps,

Jenn Visich, Ph.D.

Sr Scientist

ZymoGenetics, Inc.

1201 Eastlake Ave E

Seattle, WA 98102

(206) 442-6728

jvis.-a-.zgi.com

---

=46rom: Stephen Day

Date: Mon, 3 Dec 2001 17:29:08 -0500 (EST)

To: david.-a-.boomer.org

Subject: Re: PharmPK Defining 0.693 correction factor for converting Ke to t=

1/2

The following message was posted to: PharmPK

Richard,

David Bourne has a nice explanation here:

http://www.boomer.org/c/p1/

All you need to know is in section 4 "One Compartment

I.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 +1100

To: david.-a-.boomer.org

Subject: Re: PharmPK Defining 0.693 correction factor for converting Ke to t=

1/2

Status: R

The following message was posted to: PharmPK

Perhaps, simpler migt be:

1 Assuming a linear fall in log(concentration) v time, otherwise 1/2

lives do not exist:

Using the Naperian (natural) logs, the equation for the ln(conc), y, is y =

=3D

C(0 time) + b.time (b will be negative and is the rate constant of

elimination 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 0.693

0.693 =3D (Co + b.time1) - (Co + b.time2)

=3D b(time1 - time2)

3 1/2 life or 1/2 time =3D 0.693/b

Tom Torda

---

=46rom: "samia ezzine"

Date: Mon, 03 Dec 2001 23:03:04

To: david.aaa.boomer.org

Subject: Re: PharmPK Defining 0.693 correction factor for converting Ke to t=

1/2

Status: R

The following message was posted to: PharmPK

Hi Timothy,

As Cp =3D Coe-ket

At t1/2, Cp =3D C0/2 and t=3D t1/2 yields to C0/2 =3D C0 e-ket1/2

by simplifying e-ket1/2 =3D 1/2 , Ket1/2 =3D Ln 2 so

t1/2 =3D Ln2/ke and Ln 2=3D 0.693 so t1/2 =3D 0.693/ke

I hope that this can help you.

Samia Ezzine

ph.D student

=46acult=E9 de pharmacie

Universit=E9 de Montr=E9al

Qc, Canada

---

=46rom: "Bruce CHARLES"

Date: Tue, 4 Dec 2001 10:23:56 10

To: david.at.boomer.org

Subject: Re: PharmPK Defining 0.693 correction factor for converting Ke to t=

1/2

Status: R

The following message was posted to: PharmPK

Send reply to: PharmPK.at.boomer.org

Date 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 by

Subject: PharmPK Defining 0.693 correction factor for

converting Ke to t1/2

Going back a step further, define half-life (t1/2) as time to go from

concentration 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 / t1/2

q.e.d.

Cheers,

BC

---

From: Michael.D.Karol.aaa.abbott.com

Date: Tue, 4 Dec 2001 10:08:09 -0600

MIME-Version: 1.0

Richard:

In pharmacokinetics, we refer to half-life, the time for half of the

elimination process to be completed. The value of 0.693 is ln(2), reciproca=

l

of 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=

me

for 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. For

example, you could use the time for one third of the elimination to be

completed. This might be called a third life (as opposed to a half life). =

In

this 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 -0500

To: david.-a-.boomer.org

Subject: RE: PharmPK Defining 0.693 correction factor for converting

Ke to t1/2

The following message was posted to: PharmPK

Dr. Gendron:

0.693 comes from the natural log of 2 thus to find time, t, for half of a

concentration, Xo:

0.5Xo =3D Xo e-kt1/2

(0.5Xo/Xo)=3De-kt1/2 Xo cancels out

Ln(0.5)=3D -kt1/2

0.693=3D -kt1/2

0.693/-k =3D t1/2

Glenn Whelan, PharmD, Research Fellow

Nemours Children's Clinic

807 Children's Way

Jacksonville, FL 32250

Phone 904.390.3406

=46ax 904.390.3425

email gwhelan.-a-.nemours.org - On 5 Dec 2001 at 13:48:06, David Bourne (david.-at-.boomer.org) sent the message

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[Three more replies - db]

From: David Jaworowicz

Date: Tue, 04 Dec 2001 12:23:57 -0500

To: david.-a-.boomer.org

Subject: Re: PharmPK Defining 0.693 correction factor for converting Ke to t1/2

A more straightforward way of explaining where 0.693 comes from and why it's

used:

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 the

elimination 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 * t1/2

and thus t1/2= 0.693/k

Sincerely,

David Jaworowicz

---

From: "Daniel Sitar"

Organization: University of Manitoba

To: david.at.boomer.org

Date: Tue, 4 Dec 2001 11:44:00 -0600

Subject: Re: PharmPK Re: Defining 0.693 correction factor for

converting Ke to t1/2

Reply-to: sitar.aaa.Ms.UManitoba.CA

X-Confirm-Reading-To: sitar.at.ms.UManitoba.CA

X-pmrqc: 1

Priority: normal

Many students don't follow the math presented in these

solutions. It is better show

getting rid of the "-" sign by taking the reciprocal of e^-kt. Then

1/e^kt = 0.5

e^ kt = 2

Ln of both sides yields kt = 0.693

Dan Sitar, Pharmacology & Therapeutics

University of Manitoba

---

From: "Ronald A. Herman"

Date: Tue, 4 Dec 2001 12:54:31 -0600

To: david.-a-.boomer.org

Subject: RE: PharmPK Re: Defining 0.693 correction factor for

converting Ke to t1/2

The following message was posted to: PharmPK

As is accurately pointed out, Ln 2 = 0.693147...... so let's all stop using

0.693 and start using Ln 2. Why add rounding error? Ln 2 is on my

calculator and only takes me two keystrokes whereas, 0.693 is 5 keystrokes!

Ron Herman

Ronald A. Herman, Ph.D.

Asst. Professor (Clinical), Director

Iowa Drug Information Network

University of Iowa College of Pharmacy

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