Exponential Growth, half-life time An exponential growth function has at time $t = 5$
a) the growth factor (I guess that is just the "$\lambda$") of $0.125$ - what is the half life time?
b) A growth factor of $64$ - what is the doubling time ("Verdopplungsfaktor")?
For a), as far as I know the half life time is $\displaystyle T_{1/n} = \frac{ln(n)}{\lambda}$ but how do I use the fact that we are at $t = 5$?
I don't understand the (b) part.
Thanks
 A: The growth factor tells you the relative growth between $f(x)$ and $f(x+1)$, i.e. it's $$
  \frac{f(t+1)}{f(t)} \text{.}
$$
If $f$ grows exactly exponentially, i.e. if $$
  f(t) = \lambda\alpha^t = \lambda e^{\beta t} \quad\text{($\beta = \ln \alpha$ respectively $\alpha = e^\beta$)} \text{,}
$$
then $$
  \frac{f(t+1)}{f(t)} = \frac{\lambda\alpha^{t+1}}{\lambda\alpha^t} = \alpha = e^\beta \text{,}
$$
meaning that, as you noticed, the grow factor doesn't depend on $t$ - it's constant.
The half-life time is the time $h$ it takes to get from $f(t)$ to $f(t+h)=\frac{f(t)}{2}$. For a strictly exponential $f$, you have $$
  f(t+h) = \frac{f(t)}{2} \Rightarrow \lambda\alpha^{t+h} = \frac{\lambda}{2}\alpha^t
  \Rightarrow \alpha^h = \frac{1}{2} \Rightarrow h = \log_\alpha \frac{1}{2} = -\frac{\ln 2}{\ln \alpha} = -\frac{\ln 2}{\beta} \text{.}
$$
Similarly, the doubling-time is the time $d$ it takes to get from $f(t)$ to $f(t+d) = 2f(t)$, and you have $$
  f(t+d) = 2f(t) \Rightarrow \lambda\alpha^{t+d} = 2\lambda\alpha^t
  \Rightarrow \alpha^d = 2\Rightarrow h = \log_\alpha 2 = \frac{\ln 2}{\ln \alpha} = \frac{\ln 2}{\beta} \text{.}
$$
Thus, you always have that $d = -h$ for doubling-time $d$ and half-time $h$, which of course makes sense. If you go forward $d$ units of time to double the value, then going backwards $d$ units halves the value, similarly for going forward respectively backward $h$ units to half respectively double the value.
In your case, you get that the doubling time for (b) is $\frac{\ln 2}{\ln 64} = \frac{1}{6}$. For (a) you get that the half-life time is $-\frac{\ln 2}{\ln \frac{1}{8}} = \frac{1}{3}$.
You can also derive those by observing that a growth factor of one-eight means that going forward one unit of time means the value decreases to one-eight of the original value. Thus, after going forward one-third unit, the value decreases to one-half, since if if it decreases three times to one-half each, it overall decreases to one-eight.
Similarly, if the value increases to 64 times the original value when going forward one unit of time, you have to go forward one-sixt unit of time to have it increase to twice the value, since $2^6 = 64$.
