The enrollment capacity of a certain college is 22,200 students. If the current enrollment is 20,000, but the enrollement increases by 5% per year, how many years
will it take before the enrollment limit is reached?
We can approach this question with simple percents. Every year, the enrollment will increase by 5%.
$$20{,}000 \tag*{\tiny current enrollment}$$
$$20{,}000+20{,}000(0.05) \tag*{\tiny after 1 year}$$
We can quickly calculate 5% by first taking 10% of the number, and halving it.
In this case, 10% of 20,000 is 2,000, so 5% would be 1,000.
Therefore,
$$20{,}000(1.05)=20{,}000+1{,}000$$
$$=21{,}000$$
Repeat the process for the second year.
$$21{,}000(1.05)$$
$$=21{,}000+1{,}050$$
$$=22{,}050$$
We can estimate that another 5% increase the third year will certainly bring the number above the 22,200 threshold.
Use the compound interest equation.
$$ A=P(1+\frac{r}{n})^{nt} $$
\(A=\text{final amount}\)
\(P=\text{initial amount}\)
\(r=\text{interest rate}\)
\(n=\text{number of times interest applied for time period}\)
\(t=\text{number of time periods elapsed}\)
In our situation, we start with 20,000 and have an increase of 5% (\(r\)=0.05) once every year (\(n=1\)). We're trying to figure out what year \(t\) results in an amount greater than 22,200.
$$ 22{,}200 \lt 20{,}000\left(1+\frac{0.05}{1}\right)^{(1)t} $$
$$ 22{,}200 \lt 20{,}000(1.05)^t $$
From here we have two options for how to proceed.
We can substitute values of \(t\) and check if the inequality holds true.
For \(t=2\)
$$ 22{,}200 \lt 20{,}000(1.05)^2 $$
$$ 22{,}200 \lt 22{,}050 \ \ ✖ $$
For \(t=3\)
$$ 22{,}200 \lt 20{,}000(1.05)^3 $$
$$ 22{,}200 \lt 23{,}152.5 \ \ ✔ $$
If we know how to use logarithms, we can solve for \(t\) directly.
$$ 22{,}200 \lt 20{,}000(1.05)^t $$
$$ \frac{22{,}200}{20{,}000} \lt 1.05^t $$
$$ 1.11 \lt 1.05^t $$
$$\log_{1.05}1.11 \lt \log_{1.05}1.05^t $$
$$\frac{\log1.11}{\log1.05} \lt t $$
$$ 2.139 \lt t $$