7. Applications of Exponential Functions
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Chapter 6
7.
Applications of Exponential Functions
Exponential functions serve as powerful tools in deciphering various real-world phenomena. The lesson underlines the significance of these functions by highlighting their practical implications. For instance, they can be witnessed in areas like population growth, financial investments, and natural processes such as radioactive decay. When we commit to analyzing exponential functions, we open the door to more accurate predictions and deeper insights. The relevance of these mathematical principles becomes even more evident when we see them shaping decisions, strategies, and understandings across various domains, affirming their indispensable role in our daily experiences.
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Lesson Settings & Tools
| | 7 Theory slides |
| | 10 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Applications of Exponential Functions
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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
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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)=100b^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?
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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. cc
Linear Function & Exponential Function [0.5em]
f(x)=ax+b & g(x)=a* 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.
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, 78 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 Growingbeing 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.
Balance Afterx Months [0.4em] B(x)=850+850x 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 0.12C.
New Price of the Car C-0.12C= 0.88C 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, 78D will remain in the bloodstream.
New Concentration of the Drug: [0.5em] 7/8D 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 0.025P.
New Population Size: [0.5em] P+0.025P=1.025P 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 to 1150.
Population Afterx Years [0.5em] P(x)=1150+150x It can be concluded that this situation is modeled by a linear function. Marky's music had nothing to do with the function.
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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 15 000 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.
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)=15 000+450t
Bank B: B(m)=15 000(1+ 0.02512)^m
d Table of Values:
| Investment Period | Bank A | Bank B |
|---|---|---|
| A(t)=15 000+450t | B(m)=15 000(1+ 0.02512)^m | |
| 1 Year (12 Months) | 15 450 | ≈15 379 |
| 2 Years | 15 900 | ≈15 768 |
| 3 Years | 16 350 | ≈16 167 |
| 4 Years | 16 800 | ≈16 576 |
| 5 Years | 17 250 | ≈16 995 |
| 6 Years | 17 700 | ≈17 425 |
| 7 Years | 18 150 | ≈17 865 |
| 8 Years | 18 600 | ≈18 317 |
| 9 Years | 19 050 | ≈18 780 |
| 10 Years | 19 500 | ≈19 255 |
| 11 Years | 19 950 | ≈19 742 |
| 12 Years | 20 400 | ≈20 241 |
| 13 Years | 20 850 | ≈20 753 |
| 14 Years | 21 300 | ≈21 278 |
| 15 Years | 21 750 | ≈21 816 |
| 16 Years | 22 200 | ≈22 368 |
| 17 Years | 22 650 | ≈22 934 |
| 18 Years | 23 100 | ≈23 514 |
| 19 Years | 23 550 | ≈24 108 |
| 20 Years | 24 000 | ≈24 718 |
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.
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 15 000 v-coins, he will receive 3 % of this amount each year. Annual Profit 0.03* 15 000 = 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. Total Profit 450*10= 4500 [0.5em] Final Balance 15 000+ 4500=19 500 Therefore, Marky will be able to withdraw 19 500 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. Growth Factor 1+ 0.02512 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 120^(th) power. Use a calculator to evaluate the final balance. Final Balance [0.5em] 15 000( 1+ 0.02512)^(120)≈ 19 255 Therefore, Tiffaniqua will receive about 19 255 v-coins after ten years.
Comparison
It has been obtained that the balance of Marky's investment after 10 years will be 19 500 v-coins and Tiffaniqua's balance will be about 19 255. 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*450=9000. This will be added to his initial investment to get the final balance.
Marky's Final Balance 15 000+9000=24 000 Now, because there are 240 months in 20 years, the growth factor will be raised to 240^(th) power and multiplied by the initial investment to obtain Tiffaniqua's final balance. Tiffaniqua's Final Balance [0.5em] 15 000(1+ 0.02512)^(240)≈24 718 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 15 000 v-coins.
Balance in Bank A AftertYears A(t)=15 000+450t In case of Bank B, B(m) will be given by the product of the growth factor 1+ 0.02512 raised to the m^(th) power and the initial investment. Balance in Bank B AftermMonths [0.4em] B(m)=15 000(1+ 0.02512)^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 |
|---|---|---|
| A(t)=15 000+450t | B(m)=15 000(1+ 0.02512)^m | |
| 1 | 15 450 | ≈15 379 |
| 2 | 15 900 | ≈15 768 |
| 3 | 16 350 | ≈16 167 |
| 4 | 16 800 | ≈16 576 |
| 5 | 17 250 | ≈16 995 |
| 6 | 17 700 | ≈17 425 |
| 7 | 18 150 | ≈17 865 |
| 8 | 18 600 | ≈18 317 |
| 9 | 19 050 | ≈18 780 |
| 10 | 19 500 | ≈19 255 |
| 11 | 19 950 | ≈19 742 |
| 12 | 20 400 | ≈20 241 |
| 13 | 20 850 | ≈20 753 |
| 14 | 21 300 | ≈21 278 |
| 15 | 21 750 | ≈21 816 |
| 16 | 22 200 | ≈22 368 |
| 17 | 22 650 | ≈22 934 |
| 18 | 23 100 | ≈23 514 |
| 19 | 23 550 | ≈24 108 |
| 20 | 24 000 | ≈24 718 |
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.
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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.
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. cc Catfish & Tilapia [0.4em] y_C=55x+10 & y_T=4* 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
a
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 | 55x+10 | y_C = 55x+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* 2^x | y_T = 4* 2^x | (x,y_T) |
|---|---|---|---|
| 0 | 4*2^0 | 4 | ( 0, 4) |
| 2 | 4*2^2 | 16 | ( 2, 16) |
| 4 | 4*2^4 | 64 | ( 4, 64) |
| 6 | 4*2^6 | 256 | ( 6, 256) |
| 8 | 4*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≈ 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 = 55x + 10 | y_T = 4* 2^x |
| Growth After 6 12 Weeks | y_C = 55( 6.5)+10 | y_T = 4*2^(6.5) |
| Evaluate and Approximate | y_C≈ 368 | y_T≈ 362 |
It can be noted although x=6.5 is not the exact answer, population sizes are very close.
Catfish y_C= 55x+10 ⇒ 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. Tilapia y_T=4* 2^x ⇒ b= 2 It can be concluded that Tilapia´s population will double every week.
Extra
Applying Marky's Findings to His Fishing HopesA 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.
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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)=3x+18
Option 2: g(x)=15* 1.2^x
c Option 1: 54 000 users
Option 2: ≈ 133 700 users
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)=mx+b and an exponential function is given by the equation g(x)=a* 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)=mx+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)=mx+b ⇓ f(x)= 3x+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)=3x+b
SubstituteII
x= 0, f(x)= 18
18=3( 0)+b
ZeroPropMult
Zero Property of Multiplication
18=b
RearrangeEqn
Rearrange equation
b=18
Exponential Function
To write the function for Option 2, consider the general form of an exponential function. g(x)=a* 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* b^x ⇓ g(x)=a* 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* 1.2^x
SubstituteII
x= 0, g(x)= 15
15=a*1.2^0
▼
Solve for a
ExponentZero
a^0=1
15=a*1
IdPropMult
Identity Property of Multiplication
15=a
RearrangeEqn
Rearrange equation
a=15
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.
g(x)=15* 1.2^x
Substitute
x= 12
g(x)=15* 1.2^(12)
CalcPow
Calculate power
g(x)≈ 15(8.9)
Multiply
Multiply
g(x)≈133.5
Sells are doing great. Ignacio's family can now have an even more fantastic holiday at the theme park.
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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* 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?
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b If the aquarium contains 850 fish after 25 weeks, what is the value of b? Round to 2 decimal places.
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c Suppose that the growth factor is 2. What is the percentage growth rate per week of the fish?
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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* b^x
Substitute
x= 0
f( 0)=100* b^0
ExponentZero
a^0=1
f(0)=100*1
IdPropMult
Identity Property of Multiplication
f(0)=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* b^x
SubstituteII
f(x)= 850, x= 25
850=100* b^(25)
▼
Solve for b
DivEqn
.LHS /100.=.RHS /100.
850/100=b^(25)
CalcQuot
Calculate quotient
8.5=b^(25)
RaiseEqn
LHS^(125)=RHS^(125)
8.5^(125)=(b^(25))^(125)
PowPow
(a^m)^n=a^(m* n)
8.5^(125)=b^(25* 125)
DenomMultFracToNumber
25 * a/25= a
8.5^(125)=b^1
ExponentOne
a^1=a
8.5^(125)=b
RearrangeEqn
Rearrange equation
b=8.5^(125)
UseCalc
Use a calculator
b=1.089373...
RoundDec
Round to 2 decimal place(s)
b≈1.09
c In the given case, the fish's percentage growth rate per week will be determined if b= 2.
f(x)=100* 2^x
Substitute
x= x+1
f( x+1)=100* 2^(x+1)
▼
Rewrite
CommutativePropAdd
Commutative Property of Addition
f(x+1)=100* 2^(1+x)
SumInExponentOne
a^(1+m)=a*a^m
f(x+1)=100* 2* 2^x
CommutativePropMult
Commutative Property of Multiplication
f(x+1) = 2* 100* 2^x
Applications of Exponential Functions
Exercises
Diego has bought a very expensive designer computer for $8000. The graph represents the exponential model which describes how the computer's value depreciates over time.
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We are asked to write an exponential function with the following format. y=a* b^x In this equation, a is the initial value and b is the change factor. From the exercise, we already know that the function's initial value is the buying price a= 8000. y= 8000* b^x To determine b, we will substitute a known point on the function's graph and solve for b. We know from part A that the point (6,3000) falls on the function's graph.
Now we can complete the function. y=8000* 0.85^x
When the value of the computer decrease by 75% it will be 100 - 75 = 25% of its original worth. Multiplying the original price by 0.25 we get the decreased value.
0.25 * 8000 = 2000
Now, let's substitute 2000 for y in the equation we found in Part A and solve for t to find the time in which the computer's value reduces to 2000.
To find t, we have to use a graphing calculator. Push the Y= button and type the functions on the first two rows. Then push GRAPH to draw them. Notice that we may have to zoom in the window somewhat to see the point of intersection.
To find the point of intersection, push 2nd and CALC and choose the fifth option, intersect.
Choose the first and second curve, and pick a best guess for the point of intersection.
Given our work and the final calculations, we know that after about 8 years the computer will reduce its value by 75%.
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The following diagram shows the number of attendants for two conferences, each held annually, since 2010.
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To find the year when the conferences have approximately the same attendance, we want to find the year when the graphs intersect. From the diagram, we can see that this happens at around year 2013.
In order to estimate the exponential function which describes conference 2, we must identify two points on the graph. The first point we can mark is the function's initial value which we see is a= 12 000. This also gives us the first part of the exponential function.
y = 12 000 * b^x
To find b we must identify a second point on the curve. We have a few to choose from which are close enough to the the given lattice points. Notice that these may not be the exact coordinates, but these are acceptable since we are only looking for an approximation.
We can substitute either of these points into the function to determine b. To make the function easier to write, we will consider the year 2010 as 0 and add 1 each year that passes. Since 2015 is 5 years after 2010, the point we substitute is ( 5,6000).
Now we can write the complete function. y = 12 000 * 0.87^x
To estimate the attendance for conference 2 in the year 2030, we substitute 20 for x in the function we wrote in Part B to evaluate. Notice that we substitute x= 20 because 2030 is twenty years after 2010.
The attendance will be about 741 people if the trend continues.
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There are a total of 1024 players joining an online computer game tournament. After each round, half of the players are eliminated. How many rounds are there to reach the final match?
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Half of the players are eliminated after each round. This is means that each round we divide the number of players remaining by two. With this information, we can model the number of players during each round with an exponential function. y=ab^x At the beginning of the tournament the initial number of players is 1024. This means a= 1024. y= 1024 * b^x After each round, the amount of teams remaining is cut in half. This tells us that when x=1, we have y = 10242. Similarly, when x=2, we have y = 10242* 12. We can see that each time x increases we will multiply by 12. y=1024 * 1/2 * 1/2 * 1/2 * . . . * 1/2_(xtimes) ⇓ y=1024( 1/2)^x Now that we have our equation, we can find how many rounds will be played until the finals. Substitute 2 for y and solve for x with the Property of Equality for Exponential Equations.
The finals will begin after 9 rounds.
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You deposit $5000 in an account that earns 8 % annual interest compounded yearly. How long will it take before the account is $9000? Round to the number of years to 1 decimal place.
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Since the money grows by a percentage, we know that the account can be modeled by an exponential function. y = ab^x In this equation, a is the initial deposit which we know is $5000. This means a= 5000. Let's substitute this into the equation. y = 5000* b^x We also know that the account earns an interest of 8 %. This can be written as the change factor 1.08. Let's also substitute b= 1.08 into the function. y = 5000* 1.08^x By setting y equal to 9000, we can solve for x which tells us how many years must pass before the account has grown to $9000.
To solve this algebraically, we would need to use logarithms. This is not something we learn until Algebra 2. Therefore, we will solve this by graphing. We will write our equation as a system of equations in our graphing calculator by pressing the Y= button and typing the functions on the first two rows.
To find the point of intersection, push 2nd and CALC, then choose the fifth option intersect.
Select the first and second curve, and move to the best estimate for the point of intersection.
After about 7.6 years, the account has grown to $9000.
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At the beginning of 2022, Diego invested in a mutual fund. He has also created an equation that models the fund's expected return every quarter.y = 20 000 * 1.02^x In the model, y is the investment's value in dollars and x is the number of quarters since the beginning of 2022.
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To interpret the given formula, we will rely on the general equation of an exponential function. y = a * b^x In the general formula, a is the initial value. In the context of the exercise, the value of a represents Diego's initial investment. Given that relationship, we can say that a= 20 000. y = 20 000 * 1.02^x Therefore, Diego's initial investment was of $20 000.
In the general equation of the exponential function, b is the change factor. From the equation, we see that b= 1.02 which corresponds to a quarterly return of 2 %. y = 20 000 * 1.02^x Since there are four quarters per year, we can determine the change factor by substituting 4 for x and evaluating the power.
The annual return is given by the change factor 1.082. This is equivalent to an increase of 8.2 %.
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Dominika is tutoring a group of three students, who are studying exponential functions. To make sure her students understand applications of exponential functions, she asks them each to write a function to model the value of a zero-turn lawn mower purchased by the school's custodian for $5000!
The students write three different functions. f(x)&=5000* 1.1^x g(x)&=5000 * 0.1^x h(x)&=5000 * 0.9^x
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All three functions are written on the format of an exponential function. y=a* b^x In this equation, a is the initial value and b is the change factor. Let's analyze each function separately.
f(x)
This function has an unreasonable change factor. A change factor of 1.1 represents an appreciation in value of 10 % . Since the lawn mower will wear out it should become less valuable over time. It is not reasonable that it would increase in value.
g(x)
The function has a change factor below 1 which means it results in a depreciation. However, the change factor is 0.1 which represents a decrease of 90 % per year. For a lawn mower to drop 90 % in value in just 1 year is not reasonable.
h(x)
As with g(x), the function h(x) has a change factor that is below 1. Since the change factor is 0.9, it represents a 10 % annual decrease in value. This is reasonable for a lawn mower. Therefore, h(x) is the correct function.
In Part A, we identified h(x) as the correct function. If we substitute 3 into it, we can determine the value of the lawn mower after 3 years.
The value of the lawnmower after 3 years is $3645.
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Applications of Exponential Functions
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