Exponential behavior (growth/decay) is defined by: equal changes in the input cause the output to be multiplied by a constant .
Once you've recognized exponential behavior then
you can always use
$\,P(t) = P_{\,0}{\text{e}}^{rt}\,$ as your model.
Here, $\,P(t)\,$ is the population at time $\,t\,$, $\,P_{\,0}\,$ is the population at time zero, and $\,r\,$ is the
relative growth rate.
However, if an
(equal change,constant multiplier) = $\,(\Delta t,b)\,$ pair is readily available,
then the model $\,P(t) = P_{\,0}b^{t/\Delta t}\,$ is more natural and easier to use.
A variety of problems involving exponential functions were explored in
Solving Exponential Growth and Decay Problems.
This current section focuses on doubling time, halflife, and related problems.
Doubling time formula derived from
$\,P(t) = P_{\,0}{\text{e}}^{rt}\,$
Doubling time $\,d\,$ depends only on the relative growth rate, $\,r\,$.
Let $\,P(t) = P_{\,0}{\text{e}}^{rt}\,$ for $\,r > 0\,$ (so this is exponential growth).


As above, doubling time does not depend on time!
It depends only on the (equal change,constant multiplier) = $\,(\Delta t,b)\,$ pair.
Let $\,P(t) = P_{\,0}b^{t/\Delta t}\,$ for $\,b > 1\,$ (so this is exponential growth).
Let $\,d\,$ denote the doubling time, so that $\,P(t+d) = 2P(t)\,$:
$\,\overbrace{\cancel{P_{\,0}}b^{(t+d)/\Delta t}}^{P(t+d)} = \overbrace{2\cancel{P_{\,0}}b^{t/\Delta t}}^{2P(t)}\,$
$\,\cancel{b^{t/\Delta t}} b^{d/\Delta t} = 2\cancel{b^{t/\Delta t}}\,$
(Note that $\,t\,$ disappears at this step!)
$b^{d/\Delta t} = 2$
$\displaystyle\ln b^{d/\Delta t} = \ln 2$
$\displaystyle\frac{d}{\Delta t}\ln b = \ln 2$
$\displaystyle d = \frac{\ln 2}{\ln b}\cdot \Delta t$
Halflife is the time it takes for a current amount to be cut in half.
Halflife formula derived from
$\,P(t) = P_{\,0}{\text{e}}^{rt}\,$
Again, halflife depends only on the relative growth rate, $\,r\,$.
Let $\,P(t) = P_{\,0}{\text{e}}^{rt}\,$ for $\,r < 0\,$ (so this is exponential decay). Note that since $\,r < 0\,$ for exponential decay, we have $\displaystyle \,h = \frac{\ln 2}{r} > 0\,$. 

As above, halflife does not depend on time!
It depends only on the (equal change,constant multiplier) = $\,(\Delta t,b)\,$ pair.
Let $\,P(t) = P_{\,0}b^{t/\Delta t}\,$ for $\,0 < b < 1\,$ (so this is exponential decay).
Let $\,h\,$ denote the halflife, so that $\,P(t+h) = \frac 12P(t)\,$:
$\,\overbrace{\cancel{P_{\,0}}b^{(t+h)/\Delta t}}^{P(t+h)} = \overbrace{\frac 12\cancel{P_{\,0}}b^{t/\Delta t}}^{\frac 12P(t)}\,$
$\,\cancel{b^{t/\Delta t}} b^{h/\Delta t} = \frac 12\cancel{b^{t/\Delta t}}\,$
(Note that $\,t\,$ disappears at this step!)
$b^{h/\Delta t} = \frac 12 = 0.5$
$\displaystyle\ln b^{h/\Delta t} = \ln 0.5$
$\displaystyle\frac{h}{\Delta t}\ln b = \ln 0.5$
$\displaystyle h = \frac{\ln 0.5}{\ln b}\cdot \Delta t = \frac{\ln 2^{1}}{\ln b}\cdot \Delta t = \frac{\ln 2}{\ln b}\cdot \Delta t$
Note that since $\,0 < b < 1\,$ for exponential decay, we have $\,\ln b < 0\,$, and thus $\displaystyle h = \frac{\ln 2}{\ln b}\cdot \Delta t > 0\,$.
Let $\,g\,$ be the time it takes to go from $\,P(t)\,$ to $\,kP(t)\,$ for any $\,k > 0\,$.
The derivation is identical to those above, hence is greatly abbreviated here:
$$ \begin{gather} \cssId{s114}{P(t) = P_{\,0}{\text{e}}^{rt}}\cr\cr \cssId{s115}{P(t+g) = kP(t)}\cr\cr \cssId{s116}{P_{\,0}{\text{e}}^{r(t+g)} = kP_{\,0}{\text{e}}^{rt}}\cr\cr \cssId{s117}{{\text{e}}^{rg} = k}\cr\cr \cssId{s118}{g = \frac{\ln k}{r}} \end{gather} $$  $$ \begin{gather} \cssId{s119}{P(t) = P_{\,0}b^{t/\Delta t}}\cr\cr \cssId{s120}{P(t+g) = kP(t)}\cr\cr \cssId{s121}{P_{\,0}b^{(t+g)/\Delta t} = kP_{\,0}b^{t/\Delta t}}\cr\cr \cssId{s122}{b^{g/\Delta t} = k}\cr\cr \cssId{s123}{\frac{g}{\Delta t}\ln b = \ln k}\cr\cr \cssId{s124}{g = \frac{\ln k}{\ln b}\cdot \Delta t} \end{gather} $$ 
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
