Real Life Impacts of Analyzing Exponential Functions

Understanding the growth or decay of many real-life situations leads to making correct decisions and getting the best results. This lesson will explore situations that can be modeled using linear or exponential functions. Analyzing the rate of change will determine which of these models best fits each situation.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Challenge

Predicting Fish Growth to Win a Contest

During the holidays, Ignacio's family plans to visit the theme park Mondo Marino in California. While looking at the park's website to buy tickets, Ignacio found information about a brand new aquarium inside the park.

Wow! Ignacio notices that the park has a great contest. The person who solves a series of problems about the fish population in the aquarium will win an annual family fun pass. The website provides a function that models the fish population's growth.

f (x) = 100 b^{x}

In this function,

x

represents the weeks after the fish were introduced into the aquarium, and

b

is an unknown positive growth factor. To win the contest, these values need to be determined.

a What is the number of fish initially released into the aquarium?

b If the aquarium contains

850

fish after

25

weeks, what is the value of

b ?

Round to

2

decimal places.

c Suppose that the growth factor is

2 .

What is the percentage growth rate per week of the fish?

Example

Identifying if a Real-Life Situation Represents an Exponential or a Linear Trend

Functions can be used to model many real-life situations and solve many real-life problems. For this reason, it is essential to identify which type of function is the most appropriate for any given situation. Two of the most common types of functions are linear and exponential functions.

\underline{Linear Function} f (x) = a x + b \underline{Exponential Function} g (x) = a \cdot b^{x}

While linear functions grow by equal differences over equal intervals, exponential functions grow by equal factors. The following figure shows the graph of example linear and exponential functions.

Graph of an exponential function and a linear function

In the following situations, identify and explain whether it can be modeled by a linear or an exponential function.

a Marky just got paid after releasing his first pop single

$

Be Growing, so he opens a bank account. He deposits an initial amount of

$ 850

and plans to deposit the same quantity every month. The account earns no interest.

b The value of Marky's collection of dodgeball trading cards decreases at a rate of

12 %

per year.

c Marky takes a specific medicine to treat a fever. Every hour,

\frac{7}{8}

of the medicine's concentration in the bloodstream remains.

d A certain rabbit colony's population, started by Marky releasing his pet bunnies into the wild, increases by

2.5 %

every year.

e Marky's small town had a population of

1150

people in

1890 .

Each year, the town's population increased by

150

people. It even continued increasing despite the pop song

$

Be Growing being played non-stop on the local radio.

Answer

a Linear

b Exponential

c Exponential

d Exponential

e Linear

Hint

a The balance of the account after

x

months of the initial deposit can be determined by adding the product of

x

and the deposit to the initial amount.

b Write an expression for the new price of the dodgeball trading card collection after one year.

c Write an expression for the medicine's concentration found in the bloodstream after one hour.

d If

P

is the current population of rabbits, what will be the population after one year?

e The town's population after

x

years from

1890

can be calculated by adding the product of

x

and

150

to the initial population size.

Solution

a It is given that the person plans to deposit the quantity of

$ 850

every month. The balance

B (x)

of the account after

x

months of the initial deposit can be calculated by adding the product of

x

and

850

to the initial amount

850 .

\underline{Balance After x Months} B (x) = 850 + 850 x

This shows that the situation can be modeled by a linear function.

b The value of the dodgeball trading cards collection decreases by

12 %

each year. Let

C

represent the current price of the collection. Then, in the next year, the price of the collection will be replaced by

C

minus

0.12 C .

\underline{New Price of the Car} C - 0.12 C = 0.88 C

It can be seen that the value of the dodgeball cards will be multiplied every year by a constant factor of

0.88 .

Because this constant factor is less than

1,

it describes a decay factor. Sorry for Marky, but the given situation represents an exponential decay. This means that this is modeled by an exponential function.

c If

D

represents the current concentration of the medicine, in the next hour,

\frac{7}{8} D

will remain in the bloodstream.

\underline{New Concentration of the Drug :} \frac{7}{8} D

Similar to Part B, because this constant factor is less than

1,

this situation describes an exponential decay. This means that the situation can be modeled by an exponential function.

d Suppose that

P

represents the current size of the population of rabbits. The next year, this population will increase by

2.5 % .

Therefore, the new population size will be substituted by the sum of

P

and

0.025 P .

\underline{New Population Size :} P + 0.025 P = 1.025 P

Note that the population is multiplied every year by a constant factor of

1.025 .

Because this constant factor is greater than

1,

this is a growth factor. Therefore, this situation describes an exponential growth and is modeled by an exponential function.

e It is given that the town's population in

1890

was

1150

people. Also, each year, the population increased by

150

people. Therefore, the town's population

P (x)

after

x

years from

1890

can be calculated by adding the product of

x

and

150

1150 .

\underline{Population After x Years} P (x) = 1150 + 150 x

It can be concluded that this situation is modeled by a linear function. Marky's music had nothing to do with the function.

Example

Which Account Will Have Higher Balance at the End of the Investment Period?

Marky and his friend Tifanniqua are playing a video game in which they each have to build a city. The game gives them $15000$ v-coins for buying supplies. Since this virtual money is not enough, they must invest this initial amount. After some exploration, they have discovered two banks in the game that can help them increase their v-coins.

Marky and Tiffaniqua discussing about how to invest their v-coins

Considering that one year in the game is about $12$ minutes in the real world, the following are the offers of each virtual bank in the game.

Bank A — a simple interest of $3 %$ per year, meaning that the balance will increase by $3 %$ of the initial investment each year.
Bank B — a compound interest of $2.5 %$ per year, compounded monthly, meaning that the balance increases by one-twelfth of $2.5 %$ of the previous balance each month.

a If Marky and Tiffaniqua plan to maintain their investment for

10

years, who made the better choice?

b Instead of

10

years, they decided to keep their investment for

20

years. Does the better option change?

c Write an expression for

A (t)

representing the balance in Bank A after

t

years of the initial investment. Similarly, write an expression for

B (m)

representing the balance in Bank B after

m

months of the initial investment.

d Make a table of values showing the balances of the two banks from the first to the twentieth year. Interpret the results.

Answer

a Marky

b Tiffaniqua

c Bank A:

A (t) = 15000 + 450 t

Bank B: $B (m) = 15000 (1 + \frac{0 . 0 2 5}{1 2})^{m}$

d Table of Values:

Investment Period	Bank A	Bank B
Investment Period	$A (t) = 15000 + 450 t$	$B (m) = 15000 (1 + \frac{0 . 0 2 5}{1 2})^{m}$
$1$ Year ( $12$ Months)	$15450$	$\approx 15379$
$2$ Years	$15900$	$\approx 15768$
$3$ Years	$16350$	$\approx 16167$
$4$ Years	$16800$	$\approx 16576$
$5$ Years	$17250$	$\approx 16995$
$6$ Years	$17700$	$\approx 17425$
$7$ Years	$18150$	$\approx 17865$
$8$ Years	$18600$	$\approx 18317$
$9$ Years	$19050$	$\approx 18780$
$10$ Years	$19500$	$\approx 19255$
$11$ Years	$19950$	$\approx 19742$
$12$ Years	$20400$	$\approx 20241$
$13$ Years	$20850$	$\approx 20753$
$14$ Years	$21300$	$\approx 21278$
$15$ Years	$21750$	$\approx 21816$
$16$ Years	$22200$	$\approx 22368$
$17$ Years	$22650$	$\approx 22934$
$18$ Years	$23100$	$\approx 23514$
$19$ Years	$23550$	$\approx 24108$
$20$ Years	$24000$	$\approx 24718$

Interpretation: Bank B generates higher income after the fourteenth year. However, for the previous years, Bank B is more profitable.

Hint

a Begin by finding how much Marky will receive per year. Then determine the growth factor of Tiffaniqua's investment and compare the results.

b Using a similar reasoning as in Part A, multiply Marky's earnings per year by

20 .

For Tiffaniqua, raise the growth factor to

24 0^{th}

power and multiply it by the initial investment.

c Marky's balance after

t

years will be given by

t

times the earnings per year added to the initial investment. Furthermore, Tiffaniqua's balance after

m

months will be given by the product of the initial investment and the growth factor to the

m^{th}

power.

d Use the formulas obtained in Part C. In which year does one balance surpasses the other?

Solution

a Both Marky and Tiffaniqua plan to keep their investment for

10

years. The options will be analyzed one at a time to find the amounts Marky and Tiffaniqua will receive after

10

years.

Bank A

Bank A offers a simple interest of

3 %

per year, meaning that the balance will increase by

3 %

of the initial investment each year. Since Marky will make an initial investment of

15000

v-coins, he will receive

3 %

of this amount each year.

\underline{Annual Profit} 0.03 \cdot 15000 = 450

Bank A will give

450

v-coins to Marky for each year he keeps his investment in the bank. Now, because Marky plans to keep his investment for

10

years,

450

will be multiplied by

10

to get the total profit.

\underline{Total Profit} 450 \cdot 10 = 4500 \underline{Final Balance} 15000 + 4500 = 19500

Therefore, Marky will be able to withdraw

19500

v-coins from the bank after ten years.

Bank B

Bank B offers a compound interest of

2.5 %

per year, compounded monthly, meaning that the balance increases by one-twelfth of

2.5 %

of the previous balance each month. This situation represents an exponential growth. The growth factor is given by

1

plus the rate of growth in decimal form.

\underline{Growth Factor} 1 + \frac{0 . 0 2 5}{1 2}

The balance after

10

years, which is equal to

120

months, will be given by the product of the initial amount and the growth factor raised to

12 0^{th}

power. Use a calculator to evaluate the final balance.

\underline{Final Balance} 15000 (1 + \frac{0 . 0 2 5}{1 2})^{120} \approx 19255

Therefore, Tiffaniqua will receive about

19255

v-coins after ten years.

Comparison

It has been obtained that the balance of Marky's investment after $10$ years will be $19500$ v-coins and Tiffaniqua's balance will be about $19255 .$ Therefore, the offer of Bank A is better for the period of $10$ years.

b Following a similar reasoning as in Part A, the balance after

20

years for each investment can be determined. In this case, Marky will have a total profit of

20 \cdot 450 = 9000 .

This will be added to his initial investment to get the final balance.

\underline{Marky ’ s Final Balance} 15000 + 9000 = 24000

Now, because there are

240

months in

20

years, the growth factor will be raised to

24 0^{th}

power and multiplied by the initial investment to obtain Tiffaniqua's final balance.

\underline{Tiffaniqua ’ s Final Balance} 15000 (1 + \frac{0 . 0 2 5}{1 2})^{240} \approx 24718

It can be seen that for a

20 -

year period, the Tiffaniqua's profit will be greater. This now means that the better option is offered by Bank B.

c To find an expression for the balance

A (t)

after

t

years, the product of

t

and

450

should be added to the initial investment of

15000

v-coins.

\underline{Balance in Bank A After t Years} A (t) = 15000 + 450 t

In case of Bank B,

B (m)

will be given by the product of the growth factor

1 + \frac{0 . 0 2 5}{1 2}

raised to the

m^{th}

power and the initial investment.

\underline{Balance in Bank B After m Months} B (m) = 15000 (1 + \frac{0 . 0 2 5}{1 2})^{m}

d Using the formulas found in Part C, a table of values can be generated showing the balance of Bank A and Bank B from the first to the twentieth year. Keep in mind that

t

is equal to

12

times

m .

Investment Period in Years	Bank A	Bank B
Investment Period in Years	$A (t) = 15000 + 450 t$	$B (m) = 15000 (1 + \frac{0 . 0 2 5}{1 2})^{m}$
$1$	$15450$	$\approx 15379$
$2$	$15900$	$\approx 15768$
$3$	$16350$	$\approx 16167$
$4$	$16800$	$\approx 16576$
$5$	$17250$	$\approx 16995$
$6$	$17700$	$\approx 17425$
$7$	$18150$	$\approx 17865$
$8$	$18600$	$\approx 18317$
$9$	$19050$	$\approx 18780$
$10$	$19500$	$\approx 19255$
$11$	$19950$	$\approx 19742$
$12$	$20400$	$\approx 20241$
$13$	$20850$	$\approx 20753$
$14$	$21300$	$\approx 21278$
$15$	$21750$	$\approx 21816$
$16$	$22200$	$\approx 22368$
$17$	$22650$	$\approx 22934$
$18$	$23100$	$\approx 23514$
$19$	$23550$	$\approx 24108$
$20$	$24000$	$\approx 24718$

It can be seen that Bank B generates a higher income after the fourteenth year. However, for the previous years, Bank A is more profitable. The friends have been having such a great time playing video games. In the end, all was just fake money.

Example

Graphing Linear and Exponential Models to Identify Patterns

Besides singing, Marky is really into fishing. A small company in town called SeaBase Fish specializes in shrimp farming. They currently only farm shrimp, but they want to expand and introduce two types of fish next year, Catfish and Tilapia. The good news for Marky is that they will allow local kids the chance to fish recreationally.

Cage system on the sea where the fish is produced

Marky has never caught either type of fish in his life. He is interested in finding information about both types and then sharing this information with his friends. Online, Marky found information from another study about the growth model of each species' population after

x

weeks of the initial population being introduced into the sea.

\underline{Catfish} y_{C} = 55 x + 10 \underline{Tilapia} y_{T} = 4 \cdot 2^{x}

Marky is a bit unsure of how to model the graphs. He wonders what the growth rates of the fish will actually be.

a Graph both functions in the same coordinate plane.

b How much time after introducing the initial fish population is needed to expect that there will be the same amount of Catfish and Tilapia?

c Compare the growth rates of the two fish types.

Answer

b About six and a half weeks

c The Catfish population will increase by

55

every week. The Tilapia population will double each week.

Hint

a Begin by making a table of values for each function.

b The size of the populations will be the same when the graphs of their functions intersect.

c Linear functions grow by equal differences over equal intervals and exponential functions grow by equal factors.

Solution

a The given functions can be graphed by making a table of values. Then, each data point will be plotted in the coordinate plane. Begin with a table corresponding to Catfish's model. Because

x

is the number of weeks, only non-negative values of

x

will be used.

$x$	$55 x + 10$	$y_{C} = 55 x + 10$	$(x, y_{C})$
$0$	$55 (0) + 10$	$10$	$(0, 10)$
$2$	$55 (2) + 10$	$120$	$(2, 120)$
$4$	$55 (4) + 10$	$230$	$(4, 230)$
$6$	$55 (6) + 10$	$340$	$(6, 340)$
$8$	$55 (8) + 10$	$450$	$(8, 450)$

The data plots can be plotted and connected in a coordinate plane.

Similarly, a table for Tilapia's model will be created.

$x$	$4 \cdot 2^{x}$	$y_{T} = 4 \cdot 2^{x}$	$(x, y_{T})$
$0$	$4 \cdot 2^{0}$	$4$	$(0, 4)$
$2$	$4 \cdot 2^{2}$	$16$	$(2, 16)$
$4$	$4 \cdot 2^{4}$	$64$	$(4, 64)$
$6$	$4 \cdot 2^{6}$	$256$	$(6, 256)$
$8$	$4 \cdot 2^{8}$	$1024$	$(8, 1024)$

Now, the graph for Tilapia fish can be added by plotting and connecting the data points obtained in the second table.

b The graphs of the functions represent the population size of fish after

x

weeks they are introduced into the sea. Therefore, the

x -

value where the graphs intersect represents the number of weeks after there is approximately the same amount of Catfish and Tilapia.

It can be seen that the functions intersect at $x \approx 6.5 .$ Therefore, it is expected that after about six and a half weeks, the sizes of both populations will be approximately equal. This can be verified by evaluating each function when $x = 6.5 .$

	Catfish	Tilapia
Growth Function	$y_{C} = 55 x + 10$	$y_{T} = 4 \cdot 2^{x}$
Growth After $6 \frac{1}{2}$ Weeks	$y_{C} = 55 (6.5) + 10$	$y_{T} = 4 \cdot 2^{6.5}$
Evaluate and Approximate	$y_{C} \approx 368$	$y_{T} \approx 362$

It can be noted although $x = 6.5$ is not the exact answer, population sizes are very close.

c To compare the growth of the two populations, the rate of growth of each function needs to be identified. For the linear equation, this is given by the slope

m

of the line.

\underline{Catfish} y_{C} = 55 x + 10 \Rightarrow m = 55

This means that the Catfish population will have a constantly increase of

55

every week. Next, the growth rate of the exponential growth is the growth factor given by the base

b

of the exponent.

\underline{Tilapia} y_{T} = 4 \cdot 2^{x} \Rightarrow b = 2

It can be concluded that Tilapia´s population will double every week.

Extra

Applying Marky's Findings to His Fishing Hopes

A possible interpretation that Marky can make from his findings is that after six-and-a-half weeks after the fish are introduced, he can focus on catching Tilapia. That interpretation can be made assuming that Marky wants to catch the fish species that has such a high population that it is easier to catch.

Example

Determining User Growth on Social Media

Ignacio's family is going to have some large expenses for a holiday trip, so Ignacio's mom asked him for a hand. She wants to post some ads on social media to reach more clients to generate sales for her a client. She represents an up-and-coming musician named Marky. There is one problem, Ignacio has no idea which social media site is the best option.

For this reason, Ignacio collected some data about how the number of users of these social media sites increases on a weekly basis.

User Growth per Week on Each Social Media (in Thousands)
Time (in Weeks)	Option 1	Option 2
$0$	$18$	$15$
$1$	$21$	$18$
$2$	$24$	$21.5$
$3$	$27$	$26$
$4$	$30$	$31$
$5$	$33$	$37.5$

Ignacio wonders if this data can help him make the best decision. Find the following information to help him decide where they should post the ads.

a Using the data collected in the table, determine the trend of user growth of each social media.

b Write a function that models the data of each social media.

c Predict the number of users of each social media after

1

year. Which option should Ignacio choose to post his ads?

Answer

a Option $1$ : Linear

Option $2$ : Exponential

b Option $1$ :

f (x) = 3 x + 18

Option $2$ : $g (x) = 15 \cdot 1 . 2^{x}$

c Option $1$ :

54000

users

Option $2$ : $\approx 133700$ users

Which Option Should Ignacio Choose? Option

2

Hint

a Linear functions grow by equal differences over equal intervals and exponential functions grow by equal factors.

b A linear function has the form

f (x) = m x + b

and an exponential function is given by the equation

g (x) = a \cdot b^{x} .

c Evaluate each function at

x = 12 .

Which option is expected to attract more users?

Solution

a By finding the patterns in the data sets, the model that describes each user growth can be determined. Linear functions grow by equal differences over equal intervals and exponential functions grow by equal factors. Using this information, the pattern for Option

1

will be determined.

The number of users of Option $1$ increases by a constant of $3$ thousands every week. This data follows a linear model. Using the same process, the pattern for Option $2$ will be found.

Each week, the number of users of Option $2$ grows by a factor of about $1.2 .$ That means this data follows an exponential model.

b To write the function that describes each option, the data sets will be analyzed one at a time.

Linear Function

Consider the general form of a linear function.

f (x) = m x + b

In Part A, it was found that the number of users of Option

1

increases by

3

thousands. Since the data is given in thousands,

3

represents the slope of the function. This value will be substituted for

m

in the general form.

f (x) = m x + b ⇓ f (x) = 3 x + b

Now, the value of

b

can be determined by substituting one of the data points into the above equation and solving it for

b .

In this case,

(0, 18)

will be used.

f (x) = 3 x + b

SubstituteII

$x = 0$ , $f (x) = 18$

18 = 3 (0) + b

ZeroPropMult

Zero Property of Multiplication

18 = b

RearrangeEqn

Rearrange equation

b = 18

The function that models the data for Option

1

can be written.

f (x) = 3 x + 18

Exponential Function

To write the function for Option

2,

consider the general form of an exponential function.

g (x) = a \cdot b^{x}

In this function,

b

represents the growth factor. Because the data corresponding to Option

2

increases by a factor of about

1.2,

this value will be substituted for

b

into the general form.

g (x) = a \cdot b^{x} ⇓ g (x) = a \cdot 1.2^{x}

Similar to the linear function, this function rule will be completed using the first data point

(0, 15)

corresponding to Option

2

to determine the value of

a .

g (x) = a \cdot 1 . 2^{x}

SubstituteII

$x = 0$ , $g (x) = 15$

15 = a \cdot 1 . 2^{0}

▼

Solve for

a

ExponentZero

$a^{0} = 1$

15 = a \cdot 1

IdPropMult

Identity Property of Multiplication

15 = a

RearrangeEqn

Rearrange equation

a = 15

Finally, the function for Option

2

can be completed.

g (x) = 15 \cdot 1 . 2^{x}

c To predict the number of users of each social media after

1

year, each function will be evaluated at

x = 12 .

Begin by evaluating the function for Option

1

at this value.

f (x) = 3 x + 18

Substitute

$x = 12$

f (x) = 3 (12) + 18

Multiply

f (x) = 36 + 18

AddTerms

Add terms

f (x) = 54

There will be

54000

users using the first social media after

1

year. Next, the number of users for Option

2

will be found.

g (x) = 15 \cdot 1 . 2^{x}

Substitute

$x = 12$

g (x) = 15 \cdot 1 . 2^{12}

CalcPow

Calculate power

g (x) \approx 15 (8.9)

Multiply

g (x) \approx 133.5

The number of users of Option

2

is expected to be much greater than of Option

1 .

Also, because the user growth in Option

2

is exponential, the number of users will increase more and more. Therefore, Ignacio should choose Option

2

to post his ads. This will help him to reach more clients for Marky!

Sells are doing great. Ignacio's family can now have an even more fantastic holiday at the theme park.

Closure

Fish Population Growth: Describing the Parameters

The challenge presented at the beginning of the lesson can now be solved. The theme park's website provides a function that models fish population growth into the aquarium.

f (x) = 100 \cdot b^{x}

In this function,

x

represents the weeks after the fish was introduced into the aquarium and

b

is an unknown positive growth factor. Since Ignacio wants to win the contest, he needs to determine this information.

a What is the number of initial fish released into the aquarium?

b If the aquarium contains

850

fish after

25

weeks, what is the value of

b ?

Round to

2

decimal places.

c Suppose that the growth factor is

2 .

What is the percentage growth rate per week of the fish?

Hint

a Calculate the value of the exponential function at

x = 0 .

b Substitute

25

for

x

and

850

for

f (x) .

Then, solve the resulting equation for

b .

c Substitute

x + 1

for

x .

Then, rewrite the resulting equation in terms of

f (x) .

Solution

a Consider the given function.

f (x) = 100 \cdot b^{x}

This function models the population size of the fish after

x

weeks. Evaluate this function when

x = 0

to determine the size of the initial population.

f (x) = 100 \cdot b^{x}

Substitute

$x = 0$

f (0) = 100 \cdot b^{0}

ExponentZero

$a^{0} = 1$

f (0) = 100 \cdot 1

IdPropMult

Identity Property of Multiplication

f (0) = 100

Therefore, the initial number of fish that was released into the Aquarium is

100 .

b It is given that after

25

weeks of the release, there are

850

fish. Substitute these values into the function rule. Then, solve the resulting equation for

b .

f (x) = 100 \cdot b^{x}

SubstituteII

$f (x) = 850$ , $x = 25$

850 = 100 \cdot b^{25}

▼

Solve for

b

DivEqn

$LHS / 100 = RHS / 100$

\frac{8 5 0}{1 0 0} = b^{25}

CalcQuot

Calculate quotient

8.5 = b^{25}

RaiseEqn

$LHS^{\frac{1}{2 5}} = RHS^{\frac{1}{2 5}}$

8 . 5^{\frac{1}{2 5}} = (b^{25})^{\frac{1}{2 5}}

PowPow

$(a^{m})^{n} = a^{m \cdot n}$

8 . 5^{\frac{1}{2 5}} = b^{25 \cdot \frac{1}{2 5}}

DenomMultFracToNumber

$25 \cdot \frac{a}{2 5} = a$

8 . 5^{\frac{1}{2 5}} = b^{1}

ExponentOne

$a^{1} = a$

8 . 5^{\frac{1}{2 5}} = b

RearrangeEqn

Rearrange equation

b = 8 . 5^{\frac{1}{2 5}}

UseCalc

Use a calculator

b = 1.089373 \dots

RoundDec

Round to $2$ decimal place(s)

b \approx 1.09

c In the given case, the fish's percentage growth rate per week will be determined if

b = 2 .

f (x) = 100 \cdot b^{x} ⇓ f (x) = 100 \cdot 2^{x}

Recall that this function represents the population size of the fish at any week

x

after being introduced into the aquarium. Substitute

x + 1

for

x

into this function rule to find the population size one week later.

f (x) = 100 \cdot 2^{x}

Substitute

$x = x + 1$

f (x + 1) = 100 \cdot 2^{x + 1}

▼

Rewrite

CommutativePropAdd

Commutative Property of Addition

f (x + 1) = 100 \cdot 2^{1 + x}

SumInExponentOne

$a^{1 + m} = a \cdot a^{m}$

f (x + 1) = 100 \cdot 2 \cdot 2^{x}

CommutativePropMult

Commutative Property of Multiplication

f (x + 1) = 2 \cdot 100 \cdot 2^{x}

Note that

f (x) = 100 \cdot 2^{x}

is on the right-hand side of the obtained equation.

f (x + 1) = 2 \cdot 100 \cdot 2^{x} ⇕ f (x + 1) = 2 \cdot f (x)

This means that the fish population doubles each week. Therefore, the percentage growth rate per week is

100 % .

That is amazing. Ignacio has given this information to the organizers of the contest. He has won an annual fun pass for himself and his family.

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