A population P is initially 1000. How do you find an exponential model (growth or decay) for the population after t years if the population P is increases by 200% every 6 years?

Jul 5, 2016

$P = 1000 {\left({3}^{\frac{1}{6}}\right)}^{t}$

Explanation:

its +ve growth so the basic equation is

$P = {P}_{o} {e}^{k t}$ {where P = P(t)}

manipulating....
$\frac{P}{P} _ o = {e}^{k t}$

$\ln \left(\frac{P}{P} _ o\right) = k t$

$k = \frac{1}{t} \cdot \ln \left(\frac{P}{P} _ o\right)$

"the population P is increases by 200% every 6 years"

$k = \frac{1}{6} \cdot \ln \left(\frac{{P}_{o} \left(1 + 2\right)}{P} _ o\right)$

$= \frac{1}{6} \cdot \ln 3 = \ln {3}^{\frac{1}{6}}$

so

$P = = 1000 \left({e}^{k t}\right) = 1000 {\left({e}^{k}\right)}^{t} = 1000 {\left({e}^{\ln {3}^{\frac{1}{6}}}\right)}^{t}$

$= 1000 {\left({3}^{\frac{1}{6}}\right)}^{t}$

test

t = 0, P = 1000
t = 6, P $= 1000 {\left({3}^{\frac{1}{6}}\right)}^{6} = 3000$