- #1
Miike012
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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/%)
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/%)