Exponential Decay Help Hi I don't mean to sound like a clueless student who should already know this subject but I am really stumped on the topic of exponential growth and decay. Namely this one question: 
"When zombies finally take over, the population of the Earth will decrease exponentially. Every HOUR that goes by the human population will decrease by 5%. The population today is 6,000,000,000, find a function P that gives the population of the earth d DAYS after the beginning of the zombie takeover." 
This really puts me in between a rock and a hard place. As you can imagine I researched this question and various methods of exponential growth everywhere online, in my textbook, and in my class notes, but I still come full circle and find myself back to square one, with no clue what so ever on how this works. I would really appreciate it if someone could at least point me in the right direction like how to model a decay with exponential variables involving HOURS and DAYS as the question says. I already know my initial formula being 6,000,000,000(0.85) I just need to know the exponents used! Is it 6,000,000,000(0.85)^d/60, 6,000,000,000(0.85)^d/1? Or something like that? Please help I really want to grasp this concept!
 A: Since you are doing calculus, we will use the real mathematician's exponential function $e^x$. Measure time in hours, with $t=0$ when the exponential decay began. Then
$$P(t)=P(0)e^{-kt},\tag{1}$$
where $k$ is a constant. The initial population $P(0)$ is $6\times 10^9$.
First we find $k$. We are told that $P(1)=(0.95)P(0)$. Substituting in $(1),$ we obtain
$$(0.95)P(0)=P(0)e^{-k}.$$
Do some cancellation, and take the natural logarithm of both sides. We obtain $k=-\ln(0.95)$.
Now Equation $(1)$ can be rewritten as
$$P(t)=P(0)e^{-(-t\ln(0.95))}=e^{t\ln(0.95)}=P(0)(0.95)^t.\tag{2}$$
Note that we could have obtained the form $P(t)=P(0)(0.95)^t$ far faster. For every hour that goes by, the population gets multiplied by $0.95$.
When $d$ days have elapsed, $24d$ hours have elapsed. Set $t=24d$ in either version of $(2).$
A: The number of hours in $d$ days is $24d$, so the population multiplies by $0.95$ that many times.  Hence
$$
P(d) = 6\cdot10^9\cdot 0.95^{24d}
$$
where $d$ is the number of days.
You don't need $e$ until you talk about instantaneous rates of change, i.e. derivatives.  Then you have
$$
0.95 = e^{\ln0.95}
$$
so
$$
P(d) = 6\cdot10^9\cdot e^{24(\ln0.95)d}.
$$
