Author Topic: Logarithmic Functions (Read 3549 times) Tweet Share

d0minicz · « **Reply #15 on:** February 18, 2009, 04:36:31 pm »

The number of people, $\ {N}$ , who have a particular disease at time $\ {t}$ , is given by $\ {N} = {N}_{0}e^{kt}$

a) If the number initially is 20 000 and the number decreases by 20% each year, find:
i) the value of $\ {N}_0$
ii) the value of $\ {k}$

b) How long does it take for 5000 people to be infected?

thanks =]

Glockmeister · « **Reply #16 on:** February 18, 2009, 04:51:25 pm »

Having not done any maths in months, I'll have a crack with this one.

a) If the number initially is 20 000 and the number decreases by 20% each year, find:
i) the value of $\ {N}_0$

When 20,000 people are sick, time is equal to zero

Thus, Let $N = 20,000$ , $t = 0$

$20,000 = \ {N}_0 . e^{0}$

$20,000 = \ {N}_0$

ii) the value of $\ {k}$

$\ {N} = 20000e^{kt}$

working...

When t = 1, N = 16,000 (because each year, number of people diseased reduces by 20%)

$16000 = 20000e^{k}$

$0.8 = e^{k}$

$k = \log_{e} 0.8$

b) How long does it take for 5000 people to be infected?

$Let N = 5000$

$5000 = 20000 . e^{t\log_{e} (0.8)}$

$0.25 = e^{t\log_{e} (0.8)}$

$\log_{e} 0.25 = t\log_{e} (0.8)$

$t = \frac{\log_{e} 0.25}{\log_{e} 0.8}$

$t = 6.21 years$ (to 2 decimal points)

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Author Topic: Logarithmic Functions (Read 3549 times) Tweet Share

d0minicz

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Glockmeister

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