<PRE><B>Question 536698</B>
S(n+1)=S(n)+S(n)*.08=S(n)*(1.08)&lt;P&gt;
The students in a year (S(n+1) means students in the current year) is equal to the number of students in the previous year plus .08 times the number of students in the previous year.&lt;P&gt;
In the difference equation we have to know the number of students enrolled in a previous year.  A functional equation tells us the students in a given year without knowing the number in the previous year.&lt;P&gt;
{{{S = 2000*(1.08)^t}}} is the functional equation with S=# of students and t = number of years that have passed.&lt;P&gt;
When 0 years have passed (1990) {{{S=2000*(1.08)^0=2000*1=2000}}}&lt;P&gt;
In 2005 15 years have passed so t=15.&lt;P&gt;
{{{S = 2000*(1.08)^15=6344.34=6344}}}

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