Question

A town has a population of 5000 and grows 3.5% every year. To the nearest tenth of a year, how long will it be until the population will reach 7300?

Answer 1

Answer:

Step-by-step explanation:

This is an exponential function. In order to find the answer to the question, we need to first determine what the equation is that models this information. The standard form for an exponential function is

$y=a(b)^x$ where a is the initial value and b is the growth/decay rate. If the starting population is 5000, then

a = 5000

If the population is growing, that means that it retains 100% of the initial population and is added to by another 3.5%. So in a sense the population grows 100% + 3.5% = 103.5% or, in decimal form, 1.035. So

b = 1.035

Our function is

$y=5000(1.035)^x$ where y is the ending population and x is the number of years it takes to get to that ending population. We want to know how long, x, it will be til the population reaches 7300, y.

$7300=5000(1.035)^x$ and we need to solve for x. The only way to do that is by using logs. I'll use natural logs for this.

Begin by dividing both sides by 5000 to get

$1.46=1.035^x$ and take the natural log of both sides:

$ln(1.46)=ln(1.035)^x$

The power rule for natural logs is that we can now bring the exponent down in front of the ln to get:

$ln(1.46)=xln(1.035)$ To solve for x, we now divide both sides by ln(1.035):

$\frac{ln(1.46)}{ln(1.035)}=x$

Do that division on your calculator and get that

x = 11.0 years.

That means that 11 years after the population was 5000 it will be expected to reach 7300 (as long as the growth rate remains 3.5%)