So I recently was asked to explain half-lives since most members seem to have a misconceived view of what they are exactly. Now I'm going to be laying down some fairly advanced math here so if you don’t understand I’m sorry, but I really tried to make it as basic as I possibly could.

So to start with half-lives do not mean that there is a linear decay. However the RATE at which steroids decay is linear. What do I mean by this? Take a look at this graph.

http://www.mathexpression.com/images/graph-of-y-neg-x-4.png

This is a linear graph. This is what you guys are trying to interpret steroid half-life to be like in terms of how much testosterone you have in your body as a function of time. This is not the case. What a half-life saying is the rate of change of testosterone per half-life time is linear or decays at a standard non-exponential rate.

In mathematical differential equation terms this can be represented as dY/dT=-kY (where Y is the amount of testosterone in your body and T is time. k is our decay constant. This basically factors in how fast the material decays. Smaller half-life would have a larger decay constant.

Those d’s stand for differentials. This is basically saying an infinitesimally small slice of something. dT would represent an extremely small amount of time passing. Like .00000000000000001 half-life times. This allows us to take measurements of things that don’t have linear changes.

Anyway with our expression which we just derived (dY/dT=-kY) we can separate our variables and integrate. This is basically what a differential equation is (although this is extremely basic and the easiest form of differential equations).

Separating our variables we get

Dy/Y= (-k)dT

Now that our variables are separated we can integrate to find the amount of testosterone per unit half life. If you don’t know how to integrate from calculus I can’t really help you. It would take too long to explain. You’ll just have to trust me that I am integrating properly.

Ln(Y)=-kT+C

C is our constant of integration. It allows us to plug in different conditions without having to completely change our equation. It just represents a number that we can calculate. In this equation C will turn out to be the initial amount of testosterone injected (although this will only solve for the half life decay of a single injection, trying to explain a system of differential equations would be complex so after I finish showing half life of a single injection I’ll move onto excel and you’ll just have to trust my numbers)
Now to solve for Y (amount of testosterone in your body at any time T). To get rid of a natural log you must e both sides. e is Euler’s number (pronounced oiler) and is roughly 2.71828.

After we e both sides we get

Y=e^(-kT)+e^C

We can simplify this to

Y=C’e^(-kt)

Now we can solve for C using the initial values we were given. Say we injected ourselves with 500mg of testosterone enanthate, that would mean at time t=0 the amount of testosterone enanthate in our system would be 500mg. So

500=C’e^(-k(0))

-k times zero is zero and any exponent to the 0 power is equal to 1 so this simplifies to
500=C’

Which makes our function

Y=500e^(-kT)

With this we can solve for the amount of testosterone we have in our system as long as we have our k (which is our rate of decay constant which would vary depending on the half-life).
Knowing the half-lives we can even solve for our decay constants. If we know that test E has a half life of 10.5 days we can solve for our k by

250=500e^(-k(10.5))

Solving for k we get .066014 which makes our equation

Y=500e^(-.066014*T)

With this we can plug in any amount of time and know how much test E there is in our body after injecting 500mg of testosterone enanthate into our bodies.

So say I wanted to know how much test E was in my body after 5.25 days. That’s half the half life. So I should have ¾ of the test E still in my body right? Wrong. This is exponential decay. The greater the amount of testosterone the faster it leaves your body. Meaning the rate at which testosterone is leaving your body is constantly changing. We can find out exactly how much testosterone is in our body with our derived expression though.

Y=500e^(-.066014*5.25)

Y=353.553mg of test E which is not ¾ of 500 (which would be 375)

Anyway I don’t know if any of you will be able to understand what I just did but hopefully you can follow it enough to understand that first of all testosterone esters are constantly decaying in your body and also that it does not decay linearly over time.