$$ P=34(1.001)^{\frac{t}{4}} $$

The equation above can be used to model the population, in thousands, of a certain city \(t\) years after 2020. According to the model, the population is prediced to increase 0.1% every \(n\) months. What is the value of \(n\)?

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Through inspection, we know that when \(t=4\), the population will increase by 0.1%.

$$ P=34(1.001)^{\frac{t}{4}} $$ $$ P=34(1.001)^{\frac{4}{4}} $$ $$ P=34(1.001)^1 $$ $$ =\underbrace{34}_\text{initial population} \cdot \underbrace{(1.001)}_\text{0.1\% increase} $$

There are \(\boxed{48} \) months in 4 years.

Inspect the units of the exponent.

$$ P=34(1.001)^{\frac{t}{4}} $$ $$ \frac{t \text{ years}}{4 \text{ years}} $$

If instead we want a model which uses \(n\) months rather than \(t\) years,

$$ \frac{n \text{ months}}{4 \text{ years}} $$

The units do no cancel out, unless we convert the denominator:

$$ \frac{n \text{ months}}{4 \cancel{\text{ years}}} \cdot \frac{1 \cancel{\text{ year}}}{12 \text{ months}} $$ $$ = \frac{n \text{ months}}{ 48 \text{ months}} $$ $$ P=34(1.001)^{\frac{n}{48}} $$

Inspecting this equation, we know that when \(n=\boxed{48}\), the initial population will increase by \(0.1\%\).