# 5.04 Comparing exponential relationships

Lesson

An exponential can be represented as a graph, just like any other function. Have a look at the two plots with the four graphs below:

red: $y=1.5^x$y=1.5x; blue: $y=1.2^x$y=1.2x

The graph above shows two examples of exponential growth where both bases are bigger than $1$1, so the bigger the base, the faster the exponential grows.

red: $y=0.15^x$y=0.15x; blue: $y=0.07^x$y=0.07x

On the other hand, this graph shows two examples of exponential decay where the bases are smaller than $1$1 (and more than $0$0), so the bigger the base, the slower you decrease.

These two facts are important to know when comparing different growth and decay models so that we can say which is changing at a faster rate and which is slower.

Let's have a look at some examples to see how it applies:

#### Worked example

Amelia is considering leaving her company to start her own business. She has $40$40 current clients that will be happy to transfer over to her new company if she decides to leave. At her current position she'll gain $5$5 new clients every month, while at her new company there'll be an increase of $4%$4% every month.

(a) Form an equation for the total amount of clients ($A$A) she'll have after $t$t months for both situations.

Think: Use the formula $A=P\left(1+r\right)^t$A=P(1+r)t for her new position

Do: At her current job, she gains a total of $5\times t$5×t clients after every $t$t months, so the formula is:

$A=40+5t$A=40+5t

At her new job, it's exponential growth as the rate of $4%$4% remains the same but the amount grown increases every month. She starts out with $40$40 so the formula is:

 $A$A $=$= $40\left(1+4%\right)^t$40(1+4%)t $=$= $40\times1.04^t$40×1.04t

(b) Which job would give her more clients after $2$2 months?

Think: Use the formulae we derived, and figure out which part represents the number of months

Do:

$t$t represents the number of months so let's let $t=2$t=2:

 $A$A $=$= $40+5t$40+5t $=$= $40+5\times2$40+5×2 $=$= $40+10$40+10 $=$= $50$50
 $A$A $=$= $40\times1.04^t$40×1.04t $=$= $40\times1.04^2$40×1.042 $=$= $40\times1.0816$40×1.0816 $=$= $43.264$43.264

So she'll have more clients at her current job.

(c) In the long run, which option will give her more clients?

Think: The new position increases at an increasing rate while her current job increases at a linear rate (always the same amount)

Do:

Exponentials increase at an increasing rate so given enough time, they can always 'catch up' to linear rates and surpass them.

For example if we look at what happens $5$5 years ($60$60months) later:

 $A$A $=$= $40+5\times60$40+5×60 $=$= $40+300$40+300 $=$= $340$340
 $A$A $=$= $40\times1.04^{60}$40×1.0460 $\approx$≈ $40\times10.5196$40×10.5196 $=$= $420.784$420.784

Therefore her own business will give her more clients.

#### Practice questions

##### question 1

To investigate the environmental effect on bacterial growth, two colonies of the same bacteria were placed one in a constantly sunlit environment, the other in a dark environment. The graph shows the population of each colony after a certain number of days.

1. How many more bacteria were initially present in the sunlit environment?

2. By what percentage rate did the bacteria left in sunlight increase each day?

3. By what percentage rate did the bacteria left in darkness increase each day?

4. Initially there were more bacteria in the sunlit environment. Why then do the two curves on the graph meet?

The population of bacteria in the dark environment is decreasing at a slower rate than the population of bacteria in the sunlit environment.

A

The population of bacteria in the sunlit environment is decreasing at a slower rate than the population of bacteria in the dark environment.

B

The population of bacteria in the sunlit environment is increasing at a faster rate than the population of bacteria in the dark environment.

C

The population of bacteria in the dark environment is increasing at a faster rate than the population of bacteria in the sunlit environment.

D

The population of bacteria in the dark environment is decreasing at a slower rate than the population of bacteria in the sunlit environment.

A

The population of bacteria in the sunlit environment is decreasing at a slower rate than the population of bacteria in the dark environment.

B

The population of bacteria in the sunlit environment is increasing at a faster rate than the population of bacteria in the dark environment.

C

The population of bacteria in the dark environment is increasing at a faster rate than the population of bacteria in the sunlit environment.

D

##### question 2

Switzerland’s population in the next $10$10 years is expected to grow approximately according to the model $P=8\left(1+r\right)^t$P=8(1+r)t, where $P$P represents the population (in millions) $t$t years from now.

The world population in the next $10$10 years is expected to grow approximately according to the model $Q=7130\left(1.0133\right)^t$Q=7130(1.0133)t, where $Q$Q represents the world population (in millions) $t$t years from now.

1. According to the model, what is the current population in Switzerland?

2. According to the model, what is the current world population?

3. Switzerland’s population is expected to increase at a slower rate than the world population. Switzerland’s population increase each year could be:

$1.79%$1.79%

A

$0.87%$0.87%

B

$1.79%$1.79%

A

$0.87%$0.87%

B

### Outcomes

#### F.LE.B.5

Interpret the parameters in a linear or exponential function in terms of a context.