expoential problem

**Steph07** · December 3rd 2008, 09:17 PM

Given: There are 7 billion, that is, 7*10^9
people on Earth. If a bird flu virus starts
with one person and within 5 days two
people are infected, then each of those two
people in 5 days infects 2 people each,
and so on, how many days before everyone
on earth is infected?
Hint: Geometric series and logarithms

I know this formula....N(t)= 1e(rt) but what would the rate (r) be?
how do i do this??

**CaptainBlack** · December 3rd 2008, 11:12 PM

Originally Posted by Steph07

Given: There are 7 billion, that is, 7*10^9
people on Earth. If a bird flu virus starts
with one person and within 5 days two
people are infected, then each of those two
people in 5 days infects 2 people each,
and so on, how many days before everyone
on earth is infected?
Hint: Geometric series and logarithms

I know this formula....N(t)= 1e(rt) but what would the rate (r) be?
how do i do this??

The doubling time for number of infections is $5$ days, so after $t$ days there are $t/5$ doubling times and so the number infected is $N(t)=N(0)2^{t/5}=2^{t/5}$ (the first case occurs on day $0$ )

CB

**Steph07** · December 3rd 2008, 11:23 PM

so when I get to the ^t/5 how exactly would i solve that?? I'm sry I am a lil slow in this subject

**nzmathman** · December 4th 2008, 12:41 AM

You need to solve $2^{t/5} = 7 \times 10^9$

Take logs:
$2^{t/5} = 7 \times 10^9$

$\frac{t}{5} \times \ln(2) = \ln(7 \times 10^9)$

$t = \frac{5\ln(7\, 000\, 000\, 000)}{\ln(2)}$

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