Question 22942
{{{p(t) = 25000000e^(.015t) }}}, where t=0 in 1990.


The problem is to find p(t) in year 2000 when t=10.  In other words, find p(10).

{{{p(t) = 25000000e^(.015t) }}}
{{{p(10) = 25000000e^((.015*10)) }}}
{{{p(10) = 25000000e^(.15) }}} = 29,045,856, using a calculator.  


Since the problem is only given to 2 significant digits of accuracy, probably the answer is not more accurate than this.  Call it about 30,000,000.  I'm not sure what groundrules your instructor or the textbook may have given concerning accuracy of the problem.


For more information on this topic, especially on finding the growth constant, see my lesson plan in algebra.com in "logarithms" topic on Population Growth.


R^2 at SCC