# logistic growth

• November 19th 2010, 07:02 AM
math321
logistic growth
i really dont no what the hell to do for this question

can some1 guide me tru this 1

The logistic growth function
f(t) = (83,000 ) / 1 + 2074e^-1.6t

models the number of people who have become ill with a
particular infection t weeks after its initial outbreak in a particular community. How many people were ill after
8 weeks?
• November 19th 2010, 07:07 AM
Unknown008
$f(t) = \dfrac{83000}{1+2074e^{-1.6t}}$

Put t = 8 (Smile)
• November 19th 2010, 08:42 AM
math321
http://www.mathhelpforum.com/math-he...d0d8c99cd1.png

therefore
$
f(t) = \dfrac{83000}{1+2074e^{-1.6(8)}}

f(t) = \dfrac{83000}{1+2074e^{-12.8}}
$

AM I GOING CORRECT
• November 19th 2010, 08:44 AM
Unknown008
Yes, and the tabs are [tex] and not [code] (Giggle)
• November 19th 2010, 08:50 AM
math321
is this correct

$
f(t) = \dfrac{83000}{(1+2074)(2.76)}

f(t) = \dfrac{83000}{5727}

f(t) = 14.49
$
• November 19th 2010, 08:57 AM
Unknown008
From here... no.

$e^{-12.8} = 2.76 \times 10^{-6}$

This multiplied by 2074 will give a very small number, less than 1.
• November 19th 2010, 09:00 AM
math321
From here... no.

$e^{-12.8} = 2.76 \times 10^{-6}$

This multiplied by 2074 will give a very small number, less than 1.
• November 19th 2010, 09:06 AM
math321
$

f(t) = \dfrac{83000}{1+2074(0.0000276)}

f(t) = \dfrac{83000}{1+0.0572}

f(t) = \dfrac{83000}{1.06}
$

is this it
• November 19th 2010, 09:10 AM
Unknown008
Quite.

You missed a zero.

$f(t) = \dfrac{83000}{1+2074(0.00000276)}$

$f(t) = \dfrac{83000}{1+0.00572}$
• November 19th 2010, 09:13 AM
math321
$
f(t) = \dfrac{83000}{1.00572}
[br]
f(t) = 82527.9
$
• November 19th 2010, 09:20 AM
Unknown008
This should be okay (Smile)

When I input this in my calculator, I get:

$f(t) = \dfrac{83000}{1+2074e^{-1.6(8)}} = 82527.46077...$

It's fairly close since you too the approximation of the exponential.
• November 19th 2010, 09:36 AM
math321
thanks again for all the help u have given me