Question

On Jan 1, 2010, Chessville has a population of 50,000 people. Chessville then enters a period of population growth. Its population increases 7% each year. On the same day, Checkersville has a population of 70,000 people. Checkersville starts to experience a population decline. its population decrease 4% each year. During what year will the population Chessville first exceed that of Checkersville? Show work and explain steps.

So far I have this:

f(x) = 50000 * .07^x
f(x) = 70000 * .96^x

So how do I proceed? Do I use a table by putting x values and seeing when the population of Chessville will first exceed that of Checkersville?

Answer 1

First off, 7% growth means that each year there is 1.07 times the population. So, we need to find when

50000 * 1.07^x = 70000 * .96^x
(1.07^x/.96^x) = 70000/50000
(1.07/.96)^x = 1.4
x log(1.1146) = log(1.4)
x = log(1.4)/log(1.1146)
x = 3.101

so, after about 3 years the populations are the same

Answer 2

We aren't using logs yet. Is there any other way to do this?

Answer 3

maybe graphically.

you can see the graphs here:

http://www.wolframalpha.com/input/?i=solve+50000+*+1.07^x+%3D+70000+*+.96^x+

Hard to see how you're working with exponentials, but not yet logs.