Population Growth Problem! Helppp!?

Question

I remember doing this last year in PreCalculus and I have an assignment for AP Calculus this summer. I did almost everything except a few problems here and there, including this one! Ugh.

"The population of Silver run in year 1890 was 6,250. Assume the population increased at a rate of 2.75% a year.

a) Estimate the population in 1915 and 1940.

b) Approximately when did the population reach 50,000?"

Please be detailed so I can understand.

Answer 1

In 1891 the population would be  6250 x (1 + 0.0275)

In  1892 the population would be  [6250 x (1 + 0.0275)] x (1 + 0.0275)

In 1893 the population would be  { [6250 x (1 + 0.0275)] x (1 + 0.0275) } x (1 + 0.0275)

And so on.  This is expressed more concisely as

P  =  6250(1.0275)^t  (where t is the number of years since 1890)

So for t = 25,  P = 6250(1.0275)^25  =  6250 x 1.9704  = 12314.8

And for t = 50,  P = 6250(1.0275)^50  =  6250 x 3.8823  =  24264.5

And to reach 50000,

50000  =  6250(1.0275)^t

So  1.0275^t  =  50000 / 6250  =  8

Your calculator might do this calculation directly, or you could use logarithms :  t x log 1.0275  =  log 8

t  =  (log 8) / (log 1.0275)  =  76.6, say 77 years.

Hope that helps !

Answer 2

1915-1890 =25 years

a)6,250(1.0275)^25 is the population in 1915

=12314.755 or 12,315

b)solve 6,250(1.0275)^t = 50,000 for t

(1.0275)^t = 50,000 / 6,250 = 8

t ln (1.0275) = ln (8)

t = ln (8) / ln (1.0275) =76.65

79 years from 1890 = year 1969 it reaches 50,000