# """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 without a doubt puts me in between a rock and a difficult place. As you may consider I researched this query and numerous methods of exponential boom everywhere on line, in my textbook, and in my class notes, but I nonetheless come complete circle and locate myself lower back to square one, and not using a clue what so ever on how this works. i might really admire it if someone should at least factor me inside the right direction like the way to version a decay with exponential varia

"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 without a doubt puts me in between a rock and a difficult place. As you may consider I researched this query and numerous methods of exponential boom everywhere on line, in my textbook, and in my class notes, but I nonetheless come complete circle and locate myself lower back to square one, and not using a clue what so ever on how this works. i might really admire it if someone should at least factor me inside the right direction like the way to version a decay with exponential variables regarding HOURS and DAYS because the query says. I already recognize my preliminary method being 6,000,000,000(0.85) I simply need to know the exponents used! Is it 6,000,000,000(0.eighty five)^d/60, 6,000,000,000(zero.eighty five)^d/1? Or some thing like that? Please assist I actually need to understand this idea!
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Monserrat Ellison
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
$\begin{array}{}\text{(1)}& P\left(t\right)=P\left(0\right){e}^{-kt},\end{array}$(1)
where k is a constant. The initial population $6×{10}^{9}$.
First we find k. We are told that $P\left(1\right)=\left(0.95\right)P\left(0\right)$. Substituting in (1), we obtain
$\left(0.95\right)P\left(0\right)=P\left(0\right){e}^{-k}.$
Do some cancellation, and take the natural logarithm of both sides. We obtain $k=-\mathrm{ln}\left(0.95\right)$.
Now Equation (1) can be rewritten as
$\begin{array}{}\text{(2)}& P\left(t\right)=P\left(0\right){e}^{-\left(-t\mathrm{ln}\left(0.95\right)\right)}={e}^{t\mathrm{ln}\left(0.95\right)}=P\left(0\right)\left(0.95{\right)}^{t}.\end{array}$ (2)
Note that we could have obtained the form $P\left(t\right)=P\left(0\right)\left(0.95{\right)}^{t}$ 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).