The capacity of a school is 1100 students and the current enrollment is 980 students. If the student population increases at a rate of 5 percent per year, what is the smallest integer $n$ such that the enrollment will exceed the capacity in $n$ years?
I need to find the smallest integer $n$ that makes the inequality $980 \times 1.05^n > 1100$ true. I could try to solve this inequality algebraically, but it might be easier to use a calculator and try some values of $n$ until I find the smallest one that works. If $n = 1$, then $980 \times 1.05 = 1029$, which is not enough. If $n = 2$, then $980 \times 1.05^2 = 1080.45$, which is still not enough. If $n = 3$, then $980 \times 1.05^3 = 1134.4725$, which is more than enough. So $n = 3$ is the smallest integer that makes the inequality true. Therefore, the enrollment will exceed the capacity in 3 years. 