Question 23055
NOTE:  For more examples like this one "In Living Color" see my Lesson Plan "Population Growth" in the topic "Logarithm" in algebra.com.   


{{{P=2500e^(kt)}}}


P= 1350 in 1945 when t= -45 (45 years before 1990)
{{{1350 = 2500 e^(-45k)}}}


Divide both sides by  2500:
{{{1350/2500 = e^(-45k) }}}


Take the ln of both sides:

{{{ln(1350/2500) = ln e^(-45k) = -45k}}}


Divide both sides by -45:
{{{(ln(1350/2500))/-45= k}}}
k = .0136930253 approximately.


If you have a graphing calculator, store that value, and then use the calculator to evaluate the following, when t= 20 (years from 1990 to 2010) and k is the value from the calculator above:

{{{P= 2500 e^((20*k)) }}}
{{{P = 3287.578}}} or {{{P=3288}}} 


NOTE:  This problem can also be solved by letting the initial population be 1350 in year 1945, letting P= 2500 when t=45 in 1990.  The value of k is EXACTLY the same!  Then, to find the population in the year 2010, using the initial population of 2500, use the value of t = 20 years from 1990 to 2010.  The result is EXACTLY the same!!  Try it!!


R^2 at SCC 


P.S.  Do I get THREE points for solving this same problem THREE times???