expoential problem

• Dec 3rd 2008, 09:17 PM
Steph07
expoential problem
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??
• Dec 3rd 2008, 11:12 PM
CaptainBlack
Quote:

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

CB
• Dec 3rd 2008, 11:23 PM
Steph07
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
• Dec 4th 2008, 12:41 AM
nzmathman
You need to solve $\displaystyle 2^{t/5} = 7 \times 10^9$

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

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

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