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If the number of Ebola cases are doubling every three weeks, how long before everyone in the world ($7$ billion population) will be infected? Sorry, I forgot to add that currently we have $3000$ cases. I'm just an English major but math layman. I think this may be too scary for many of us to consider. My friend offered this: $2^n=7000000000=7*10^9$ Solve for $n$ Then $3n=$number of weeks

Stefan Hamcke
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Ed S.
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    How many are infected by ebola at the start? – Alice Ryhl Sep 25 '14 at 15:28
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    I don't see why this question is getting downvoted. Why? Because its easy? I think we need to get away from the idea that the difficulty of a question is somehow a measure of its legitimacy. – goblin GONE Sep 25 '14 at 15:43
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    @goblin: Indeed it is our duty as mathematicians to explain how answers to such problems are arrived at, and to point out inadequacies of the model (if the number of Ebola cases). – André Nicolas Sep 25 '14 at 16:15
  • @AndréNicolas, yep. Agreed 100%. Realistically, these sorts of exponential models always ignore factors that ultimately prevent the exponential growth. A silly example: if more than half the world is infected with Ebola, then the number of infected people can hardly double in 3 weeks. – goblin GONE Sep 26 '14 at 04:30

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Assume that in the beginning there are $100$ cases which double every three weeks, then we have $$f(n)=100 \cdot 2^{n/3}$$ which is the function that gives the number of cases after $n$ weeks have passed. To see the logic behind it, note that if you plug in $n=3$ weeks, it should double and $f(3)=200$ as it should be. I hope the logic behind the function is clear, ask if it's not.

Now to find the number of weeks till the cases are $7 \cdot 10^9$ we let $f(n)=7 \cdot 10^9$ and solve for $n$:

$$100 \cdot 2^{n/3}=7\cdot 10^9 \implies 2^{n/3} = 7 \cdot 10^7 \implies \frac n3 \log 2 = \log (7 \cdot 10^7) \implies n = 3\frac{\log (7\cdot 10^7)}{\log 2} \approx 78 \text{ weeks}$$

A dangerously low number, only a year and a half. That is why it is such an important international concern.

Sheheryar Zaidi
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  • Your last paragraph is misleading, since this is not meant to be anything like a realistic model. – Kevin Carlson Sep 26 '14 at 05:04
  • @KevinCarlson True, I am not sure about the exact figures given by the OP, but it is true that such models are exponential, which in turn shows how quickly the number of cases can grow. Doesn't that justify the last sentence? – Sheheryar Zaidi Sep 26 '14 at 05:06