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Population y grows according to the equation dy/dx=ky, where k is constant and t is measured in years. If the population doubles every 10 years, then what is the value of k?

1 Answer

Mar 4, 2018

$K = \frac{ln 2}{10}$ $K=ln2/10$

Explanation:

Standard equation for 'the law of natural growth' is $P [t] = C e^{k t}$ $P[t]=Ce^[kt$ .

Let $P [t] = 1$ $P[t]=1$ when $t = 0$ $t=0$ , and so $1 = C e^{k [0],}$ $1=Ce^[k[0],$ ie $C = 1$ $C=1$

So when the population has doubled to $2$ $2$ ,

$2 = e^{10 k}$ $2=e^[10k$ , taking logs of of both sides ln $2 = ln [e^{10 k}$ $2=ln[e^[10k$ ]

ln $2 = 10 k$ $2=10k$ [Theory of logs] therefore $k = \frac{ln 2}{10}$ $k=ln2/[10]$ . Hope this was helpful.

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