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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| | 7 Theory slides |
| | 10 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Applications of Exponential Functions
Slide of 7
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.
f(x)=100bx
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. 
In the following situations, identify and explain whether it can be modeled by a linear or an exponential function.
Linear Functionf(x)=ax+bExponential Functiong(x)=a⋅bx
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.
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, 87 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 After x MonthsB(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 CarC−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, 87D will remain in the bloodstream.
New Concentration of the Drug:87D
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: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 After x YearsP(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. 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.
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+450t
Bank B: B(m)=15000(1+120.025)m
d Table of Values:
| Investment Period | Bank A | Bank B |
|---|---|---|
| A(t)=15000+450t | B(m)=15000(1+120.025)m | |
| 1 Year (12 Months) | 15450 | ≈15379 |
| 2 Years | 15900 | ≈15768 |
| 3 Years | 16350 | ≈16167 |
| 4 Years | 16800 | ≈16576 |
| 5 Years | 17250 | ≈16995 |
| 6 Years | 17700 | ≈17425 |
| 7 Years | 18150 | ≈17865 |
| 8 Years | 18600 | ≈18317 |
| 9 Years | 19050 | ≈18780 |
| 10 Years | 19500 | ≈19255 |
| 11 Years | 19950 | ≈19742 |
| 12 Years | 20400 | ≈20241 |
| 13 Years | 20850 | ≈20753 |
| 14 Years | 21300 | ≈21278 |
| 15 Years | 21750 | ≈21816 |
| 16 Years | 22200 | ≈22368 |
| 17 Years | 22650 | ≈22934 |
| 18 Years | 23100 | ≈23514 |
| 19 Years | 23550 | ≈24108 |
| 20 Years | 24000 | ≈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.
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 mth 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.Annual Profit0.03⋅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. Total Profit450⋅10=4500Final Balance15000+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.Growth Factor1+120.025
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 120th power. Use a calculator to evaluate the final balance.
Final Balance15000(1+120.025)120≈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⋅450=9000. This will be added to his initial investment to get the final balance.
Marky’s Final Balance15000+9000=24000
Now, because there are 240 months in 20 years, the growth factor will be raised to 240th power and multiplied by the initial investment to obtain Tiffaniqua's final balance. Tiffaniqua’s Final Balance15000(1+120.025)240≈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.
Balance in Bank A After t YearsA(t)=15000+450t
In case of Bank B, B(m) will be given by the product of the growth factor 1+120.025 raised to the mth power and the initial investment.
Balance in Bank B After m MonthsB(m)=15000(1+120.025)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)=15000+450t | B(m)=15000(1+120.025)m | |
| 1 | 15450 | ≈15379 |
| 2 | 15900 | ≈15768 |
| 3 | 16350 | ≈16167 |
| 4 | 16800 | ≈16576 |
| 5 | 17250 | ≈16995 |
| 6 | 17700 | ≈17425 |
| 7 | 18150 | ≈17865 |
| 8 | 18600 | ≈18317 |
| 9 | 19050 | ≈18780 |
| 10 | 19500 | ≈19255 |
| 11 | 19950 | ≈19742 |
| 12 | 20400 | ≈20241 |
| 13 | 20850 | ≈20753 |
| 14 | 21300 | ≈21278 |
| 15 | 21750 | ≈21816 |
| 16 | 22200 | ≈22368 |
| 17 | 22650 | ≈22934 |
| 18 | 23100 | ≈23514 |
| 19 | 23550 | ≈24108 |
| 20 | 24000 | ≈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.
CatfishyC=55x+10Tilapia yT=4⋅2x
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 | yC=55x+10 | (x,yC) |
|---|---|---|---|
| 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⋅2x | yT=4⋅2x | (x,yT) |
|---|---|---|---|
| 0 | 4⋅20 | 4 | (0,4) |
| 2 | 4⋅22 | 16 | (2,16) |
| 4 | 4⋅24 | 64 | (4,64) |
| 6 | 4⋅26 | 256 | (6,256) |
| 8 | 4⋅28 | 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 | yC=55x+10 | yT=4⋅2x |
| Growth After 621 Weeks | yC=55(6.5)+10 | yT=4⋅26.5 |
| Evaluate and Approximate | yC≈368 | yT≈362 |
It can be noted although x=6.5 is not the exact answer, population sizes are very close.
CatfishyC=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.
TilapiayT=4⋅2x⇒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.
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)=3x+18
Option 2: g(x)=15⋅1.2x
c Option 1: 54000 users
Option 2: ≈133700 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⋅bx.
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
f(x)=3x+18
Exponential Function
To write the function for Option 2, consider the general form of an exponential function.g(x)=a⋅bx
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⋅bx⇓g(x)=a⋅1.2x
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⋅1.2x
SubstituteII
x=0, g(x)=15
15=a⋅1.20
▼
Solve for a
ExponentZero
a0=1
15=a⋅1
IdPropMult
Identity Property of Multiplication
15=a
RearrangeEqn
Rearrange equation
a=15
g(x)=15⋅1.2x
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⋅1.2x
Substitute
x=12
g(x)=15⋅1.212
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.
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⋅bx
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⋅bx
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⋅bx
Substitute
x=0
f(0)=100⋅b0
ExponentZero
a0=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⋅bx
SubstituteII
f(x)=850, x=25
850=100⋅b25
▼
Solve for b
DivEqn
LHS/100=RHS/100
100850=b25
CalcQuot
Calculate quotient
8.5=b25
RaiseEqn
LHS251=RHS251
8.5251=(b25)251
PowPow
(am)n=am⋅n
8.5251=b25⋅251
DenomMultFracToNumber
25⋅25a=a
8.5251=b1
ExponentOne
a1=a
8.5251=b
RearrangeEqn
Rearrange equation
b=8.5251
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⋅bx⇓f(x)=100⋅2x
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⋅2x
Substitute
x=x+1
f(x+1)=100⋅2x+1
▼
Rewrite
CommutativePropAdd
Commutative Property of Addition
f(x+1)=100⋅21+x
SumInExponentOne
a1+m=a⋅am
f(x+1)=100⋅2⋅2x
CommutativePropMult
Commutative Property of Multiplication
f(x+1)=2⋅100⋅2x
f(x+1)=2⋅100⋅2x⇕f(x+1)=2⋅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.
Applications of Exponential Functions
Exercises
First Month: increase by 5%Second Month: decrease by 3%
Dylan calculated that when his bank account has grown by 25%, he will have just enough money to buy the console.{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.80556em;vertical-align:-0.05556em;\"><\/span><span class=\"mspace\" style=\"margin-right:0.16666666666666666em;\"><\/span><span class=\"mord\">%<\/span><\/span><\/span><\/span>","answer":{"text":["11.6"]}}
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An increase of 5 % means that the resulting quantity is 105% of the original quantity. This equals to a change factor of 1.05. In a similar manner, decrease of 3 % equals a change factor of 1 - 0.03 = 0.97. Now we can distinguish that there are two different factors for each month. First Month:& 1.05 Second Month:& 0.97 During this period of two months, Dylan's bank account undergoes both of these changes. The combined change factor is the product of 1.05 and 0.97. 1.05* 0.97=1.0 185 If his bank account contains a dollars at the beginning of the year, it will contain 1.85 % more dollars at the end of this two month period. If we consider that this pattern repeats itself, we get the following exponential function. y = a* 1.0185^x In this equation, x represents the number of two month intervals since Dylan started saving. Notice that in one year we have 6 two month periods. Therefore, we must substitute 6 for x in this equation and evaluate.
The combined change factor after one year is 1.116. This corresponds to an increase of 11.6 %.
We want to know when the bank account is 25 % greater than it was at the beginning of the year. This is 125% of Dylan's initial savings a. So, we want to determine when y= 1.25a.
To find x, 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 out 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.
After about 12.17 of these two-month periods Dylan has saved up enough money. This equals a total time of 12.17* 2= 24.34. We round this up to 25 months. If Dylan wants to buy the console in a year, he needs to review his expenses closely.
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Applications of Exponential Functions
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