Question

Exponential Growth. The population of a city 10 years ago was 45,600. Since then the population has increased at a steady rate each year. If the population is currently 64,800, find the annual rate of growth for this city. Please show work! For this problem the answer is 3.6%.

Answer 1

i think you are supposed to solve
\[64800=45600e^{10k}\] for k

Answer 2

let's see if it works:
\[\frac{27}{19}=e^{10k}\]
\[\ln(\frac{27}{19})=10k\]
\[k=\frac{\ln(\frac{27}{19})}{10}=.035\]\

Answer 3

close
was this the method, or where you supposed to do something else?

Answer 4

Since the rate is steady I think you should do Y=a(1+r)^t

Answer 5

exponential growth rate is very steady, but we can try it this way too

Answer 6

I figured it out

Answer 7

\[64,800=54,600(1+r)^{10}\] and solve for \(r\)
\[\frac{27}{19}=(1+r)^{10}\]
\[1+r=\sqrt[10]{\frac{27}{19}}\]
\[1+r=1.0358\]
\[r=.0358\approx 3.6\%\]