1. ## Exponents word problem...

Since January 1980, the population of the city of Brownville has grown according to the mathematical model y=720,500(1.022)^x, where x is the number of years since January 1980. Explain what the numbers 720,500 and 1.022 represent in this model. If this trend continues, use this model to predict the year during which the population of Brownville will reach 1,548,800.

How do i figure this out?

2. Hello, eh501!

Since January 1980, the population of the city of Brownville has grown
according to the mathematical model: $y\:=\:720,\!500(1.022)^x$
where $x$ is the number of years since January 1980.

(a) Explain what the numbers 720,500 and 1.022 represent in this model.

Let $x=0$ and we have: . $y \:=\:720,\!500(1.022)^0 \:=\:720,\!500$

That is, when time is 0, the population was 720,500.

Therefore, 720,500 is the initial population of the city (in 1980).

The population is growing by a factor of 1.022 each year.
This means that the population is growing by 2.2% annually.

(b) If this trend continues, use this model to predict the year
during which the population of Brownville will reach 1,548,800.

Then question is: for what $x$ will $y = 1,\!548,\!000$ ?

We have: . $720,\!500(1.022)^x \:=\:1,\!548,\!000 \quad\Rightarrow\quad (1.022)^x \:=\:\frac{1,\!548,\!800}{720,\!500} \:=\:\frac{1408}{655}$

Take logs: . $\log(1.022)^x \:=\:\log\left(\frac{1408}{655}\right) \quad\Rightarrow\quad x\cdot\log(1.022) \:=\:\log\left(\frac{1408}{655}\right)$

Hence: . $x \;=\;\frac{\log(\frac{1408}{655})}{\log(1.02)} \;=\;35.16718012$

Therefore, the population will reach 1,548,800 during the 35th year (2015).