Question 750242
Simplest way to analyze it is that each year population is pop*(1.00-0.12).


Starting with 125 students,
1 year 125*(1-0.12)
2 year {{{125*(1-0.12)^2}}}
3 year {{{125*(1-0.12)^3}}}
n year {{{highlight(125*(1-0.12)^n=125*(0.88)^n)}}}


You are looking for how {{{75=125(0.88)^n}}}
then
{{{highlight(0.6=(0.88)^n)}}}


Can you take this the rest of the way to the solution for n? Take logarithms of both sides, and use the rule about logs,  {{{log((b^a))=a*log((b))}}}