+956 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
+9479 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 
+9479 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
