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

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 $0.1\%$.