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
**+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 = 1.06^{n}

ln(70/36.5) = ln(1.06^{n})

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

#1
**+4 **

Best Answer

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 = 1.06^{n}

ln(70/36.5) = ln(1.06^{n})

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