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The growth in population of a town since 2000 is given, in thousands, by the function P(n) = 36.5(1.06)n. In which year will the population expect to reach 70 000?

2011

2010

2013

2008

Julius  Apr 20, 2017

#1
+4155
+4

I can try, but I might get it wrong!

Is this the function: P(n) = 36.5 (1.06)n

We want to solve for n when P(n) = 70

70 = 36.5 (1.06)n

70/36.5 = 1.06n

ln(70/36.5) = ln(1.06n)

ln(70/36.5) = n * ln(1.06)

$$\frac{\ln(70/36.5)}{\ln(1.06)}=n\approx11.176$$

So..since the function considers the year 2000 the starting point,

The population should reach 70 thousand in the year 2011

hectictar  Apr 20, 2017
edited by hectictar  Apr 20, 2017
Sort:

#1
+4155
+4

I can try, but I might get it wrong!

Is this the function: P(n) = 36.5 (1.06)n

We want to solve for n when P(n) = 70

70 = 36.5 (1.06)n

70/36.5 = 1.06n

ln(70/36.5) = ln(1.06n)

ln(70/36.5) = n * ln(1.06)

$$\frac{\ln(70/36.5)}{\ln(1.06)}=n\approx11.176$$

So..since the function considers the year 2000 the starting point,

The population should reach 70 thousand in the year 2011

hectictar  Apr 20, 2017
edited by hectictar  Apr 20, 2017

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