Calculate Half Life Values using First Order Kinetics Formula

[Miike012]

Formula consists of calculating half lifes of a first order... T (1/2) = 0.693/k

1. T (1/2)(x) = T (%)

I noticed at various half life values such as T(1/2) , T(1/4) , T(1/8) ... T(%)
That these values are the product of T(1/2) and some Factor "X"... As I stated in formula 1.

-Next I noticed that at every half life each factor of X increased by one
Example: T(1/2) X=1 , T(1/4) X=2, T(1/8) X=3 ...

-Thus if I know the percentage of the half life I can work backwards. Thus determing the Factor X
Example: T(3.125%)

3.125/100 = 1/x ----> x = 32
Thus: T(3.125%) = T(1/32)
At T(1/32) the factor of X is 5... X = 5
And T(1/2)(5) = T(1/32)

-Solving for X
1/32 = 1/2^X ----> 32 = 2^X -----> X = Log32/Log2

*Remeber number 32 is a percentage thus 32 can be substituted for 100/%
and X = Log(100/%)/Log2 ... The base will be a constant equal to 2.

-Finally: Taking both equations and substituting:

1. T(1/2)(X) = T(%)
2. X = Log(100/%)/Log2

Derived formula: T(1/2)*Log(100/%)/Log2 = T(100/%)

 

[Miike012]

Anyone ??

 

[LCKurtz]

Perhaps the reason you have no responses is that your notation is poor and it isn't clear what you are trying to accomplish. If you want to know the value of t for any portion λ of the original amount you just solve

$$\lambda P_0 =P_0e^{-kt}$$
for t:
$$t = -k\ln(\lambda)$$
If λ = 1/2ⁿ this gives (denoting this t as t_n)
$$t_n=-k\ln(\frac 1 {2^n}) =kn\ln(2)= nt_1$$
where t₁ is the time for the first half life.

I guess that's what you are getting at, but I'm not sure what it gets you. And I don't see the point of using percentages instead of proportions.