"""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

Hrefnui9 2022-09-14 Answered
"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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Answers (1)

Monserrat Ellison
Answered 2022-09-15 Author has 22 answers
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
(1) P ( t ) = P ( 0 ) e k t , (1)
where k is a constant. The initial population 6 × 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
(2) P ( t ) = P ( 0 ) e ( t ln ( 0.95 ) ) = e t ln ( 0.95 ) = P ( 0 ) ( 0.95 ) t . (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, 24 d hours have elapsed. Set t = 24 d in either version of (2).

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