Help! Calculus question

Question

In 1960 there were 238,950 immigrants admitted to a country. In 2009 the number was 1,073,319. Write an equation expressing immigrants y, in terms of t, the number of years after 1900

Answer 1

This is an exponential growth question.  I know it was asked a little while ago and I just saw it. 

 

Basically what you know is

 

in year 0 there was 238,950 immigrants in a country

 

We set y=238,950 at time = 0

 

So we know this:

 

238,950=Ce^k(0)

e is to the 0th power so that is equal to one.  We can say the constant is 238950 or C=238950

 

now we say 49 years later the population is 1073319

 

so we are trying to solve for k to get an equation

 

so We say

 

1073319 = 238950e^k(49)

 

We will divide 1073319 by 238950

 

and get 4.491=e^49k

 

Take ln of both sides to get rid of e

 

1.502=49k then solve for k to get

 

k=.03065

 

So the general equation for your problem is

 

y=238950e^.03065t