Applications Of Exponential Functions Quiz

1. Identify the initial value here: y = 6.87 (4)^x

A=3

A = 6.87

A=4

A=1

The initial value in this equation is represented by the variable "a" and it is equal to 6.87. This means that when x is equal to 0, the value of y is equal to 6.87.

Explanation

The initial value in this equation is represented by the variable "a" and it is equal to 6.87. This means that when x is equal to 0, the value of y is equal to 6.87.

2. A gas costs $1.99 per gallon. If the price per gallon increases an average of 6% per month, which of the following function models the exponential growth of the pricing?

1.99(1.06)t

1.06(1.99)t

1.99(.94)t

[1.06(1.99)]t

The correct answer is 1.99(1.06)t. This function models the exponential growth of the pricing because it takes the initial cost of $1.99 per gallon and multiplies it by 1.06, which represents a 6% increase, raised to the power of t, which represents the number of months. This formula allows for the calculation of the increasing price over time due to the monthly growth rate.

Explanation

The correct answer is 1.99(1.06)t. This function models the exponential growth of the pricing because it takes the initial cost of $1.99 per gallon and multiplies it by 1.06, which represents a 6% increase, raised to the power of t, which represents the number of months. This formula allows for the calculation of the increasing price over time due to the monthly growth rate.

3. Which of these functions shows a starting amount of $15 and an increase of 35% every year?

Y = 35(1+ 0.15)x

Y = 15(35)x

Y = 15(1 + 0.35)x

Y = 15(0.35)x

The correct answer is y = 15(1 + 0.35)x. This equation represents a starting amount of $15 (the constant term) and an increase of 35% every year (represented by multiplying the starting amount by 1 + 0.35). The exponent x indicates the number of years.

Explanation

The correct answer is y = 15(1 + 0.35)x. This equation represents a starting amount of $15 (the constant term) and an increase of 35% every year (represented by multiplying the starting amount by 1 + 0.35). The exponent x indicates the number of years.

4. James decides to invest $20 in stock at Target. The stock raises 1.8% each year. Write an equation to model the situation then find how much money this stock will be worth in 6 years.

Y = 20 (0.018)x
$1.11

Y = 20 (0.982)x$17.93

Y = 1.8 (20)x$115,200,000

Y = 20 (1.018)x$22.26

The equation y = 20 (1.018)x models the situation because it represents the initial investment of $20 growing by 1.8% each year for x years. Plugging in x = 6, we can calculate the value of y, which is the amount of money the stock will be worth in 6 years. The value of y turns out to be $22.26.

Explanation

The equation y = 20 (1.018)x models the situation because it represents the initial investment of $20 growing by 1.8% each year for x years. Plugging in x = 6, we can calculate the value of y, which is the amount of money the stock will be worth in 6 years. The value of y turns out to be $22.26.

5. Jessica deposited $10 in a savings account earning 5% interest, compounded annually. Write an equation then find to the nearest cent, how much interest will she earn in 3 years?

Y = 10 ( 5)x$1, 250

Y = ( 1.05)x$1.15

Y = 10 ( 3)x$2, 430

Y = 10 ( 1.05)x$11.58

Jessica deposited $10 in a savings account earning 5% interest, compounded annually. The equation that represents the amount of money she will have after a certain number of years is y = 10(1.05)^x, where x is the number of years. In this case, she wants to find the interest earned in 3 years, so x = 3. Plugging in the values, we get y = 10(1.05)^3 = $11.58. Therefore, Jessica will earn $11.58 in interest in 3 years.

Explanation

Jessica deposited $10 in a savings account earning 5% interest, compounded annually. The equation that represents the amount of money she will have after a certain number of years is y = 10(1.05)^x, where x is the number of years. In this case, she wants to find the interest earned in 3 years, so x = 3. Plugging in the values, we get y = 10(1.05)^3 = $11.58. Therefore, Jessica will earn $11.58 in interest in 3 years.

6. What does the following equation represent, exponential growth or decay?
y = 5 ( 0.3)^x

Decay because 0 < 0.3 < 1

Decay because 0.3 < 1

Growth because 5 >1

None of the above

The given equation represents exponential decay because the value of the base, 0.3, is between 0 and 1. In exponential decay, the value decreases over time as the exponent increases.

Explanation

The given equation represents exponential decay because the value of the base, 0.3, is between 0 and 1. In exponential decay, the value decreases over time as the exponent increases.

7. What is the starting term, for this function: f(x) = 300(1.16)^x?

X

.16

1.16

300

The starting term for the function f(x) = 300(1.16)x is 300. This is because the function is in the form of y = a(b)x, where a is the starting term or initial value. In this case, 300 is the initial value or starting term of the function.

Explanation

The starting term for the function f(x) = 300(1.16)x is 300. This is because the function is in the form of y = a(b)x, where a is the starting term or initial value. In this case, 300 is the initial value or starting term of the function.

8. Pick the equation which is exponential decay.

Y=2000(0.88)

Y=0.5(1.88)

Both

None

The equation y=2000(0.88) represents exponential decay because the value inside the parentheses, 0.88, is less than 1. In exponential decay, the value decreases over time or with each iteration. In this equation, the initial value is 2000 and it decreases by a factor of 0.88.

Explanation

The equation y=2000(0.88) represents exponential decay because the value inside the parentheses, 0.88, is less than 1. In exponential decay, the value decreases over time or with each iteration. In this equation, the initial value is 2000 and it decreases by a factor of 0.88.

9. A child asks his dad for a pocket money that starts with a penny and then doubles every day for a month. Which of the following functions can be used to model the amount of money (A) the child will receive each day(x)?

A(x) = 0.01(2)x

A(x) = 2(0.01)x

A(x) = 0.01(1 - 2)x

A(x) = 2(1.01)x

The correct answer is A(x) = 0.01(2)x. This function represents the amount of money the child will receive each day. The function starts with 0.01, which is the initial amount of money (a penny), and then doubles every day (2x). The variable x represents the number of days.

Explanation

The correct answer is A(x) = 0.01(2)x. This function represents the amount of money the child will receive each day. The function starts with 0.01, which is the initial amount of money (a penny), and then doubles every day (2x). The variable x represents the number of days.

10. Suppose a culture of bacteria starts with 5000 cells and dies by 30% every year. Write an equation to represent this situation.

Y=30(5000)x

Y=5000(1.3)x

Y=5000xx

Y=5000(0.7)x

The equation y=5000(0.7)x represents the situation where the number of bacteria cells, y, decreases by 30% every year. The initial number of cells, 5000, is multiplied by 0.7 (which is equivalent to 1 - 0.3) to represent the decrease of 30%. The exponent x represents the number of years.

Explanation

The equation y=5000(0.7)x represents the situation where the number of bacteria cells, y, decreases by 30% every year. The initial number of cells, 5000, is multiplied by 0.7 (which is equivalent to 1 - 0.3) to represent the decrease of 30%. The exponent x represents the number of years.

11. Pick an equation that is exponential growth.

Y=2(0.5)x

Y = 0.5(2)x

Both

None

The equation y = 0.5(2)x represents exponential growth because the base, 2, is greater than 1. In exponential growth, the value of y increases rapidly as x increases. In this equation, as x increases, the value of 2x also increases, resulting in a larger value for y. The coefficient 0.5 determines the initial value or starting point of the growth.

Explanation

The equation y = 0.5(2)x represents exponential growth because the base, 2, is greater than 1. In exponential growth, the value of y increases rapidly as x increases. In this equation, as x increases, the value of 2x also increases, resulting in a larger value for y. The coefficient 0.5 determines the initial value or starting point of the growth.

12. A town doubles its size every 5 years. If the population of the town is currently 500, write an equation and find what will the population be in 20 years?

Y =5 (2)x80 people

Y = 500 (2)x8,000 people

Y = 500 (.2)x0.8 people

Y = 500 (2)x524, 288, 000 people

The equation y = 500 (2)x represents the population of the town after x number of 5-year periods. In this case, we want to find the population after 20 years, so x = 20. Plugging this into the equation, we get y = 500 (2)^20 = 8,000. Therefore, the population of the town will be 8,000 people in 20 years.

Explanation

The equation y = 500 (2)x represents the population of the town after x number of 5-year periods. In this case, we want to find the population after 20 years, so x = 20. Plugging this into the equation, we get y = 500 (2)^20 = 8,000. Therefore, the population of the town will be 8,000 people in 20 years.

13. Suppose an automobile that originally costs $14,000 depreciates by 20% of its value every year. Which exponential equation is suitable to model this situation?

Y=14000(.20)x

Y=14000(1+.80)x

Y=14000(.80)x

Y=14000(1+.20)x

The correct answer is y=14000(.80)x. This equation represents the depreciation of the automobile over time. The initial value of $14,000 is multiplied by 0.80 (which represents a 20% decrease) raised to the power of x (the number of years). This equation accurately models the situation where the automobile depreciates by 20% of its value every year.

Explanation

The correct answer is y=14000(.80)x. This equation represents the depreciation of the automobile over time. The initial value of $14,000 is multiplied by 0.80 (which represents a 20% decrease) raised to the power of x (the number of years). This equation accurately models the situation where the automobile depreciates by 20% of its value every year.

14. An equation in the form of y=b^x will represent decay if

B > 1

B < 0

0 < b < 1

B>0

An equation in the form of y=bx represents decay if the value of b is between 0 and 1. In this range, as x increases, y decreases, indicating a decay or decrease in the value of y. If b is greater than 1, the equation represents growth rather than decay. Similarly, if b is negative, the equation represents a reflection or flip of the graph, but not necessarily decay. Therefore, the correct answer is 0

Explanation

An equation in the form of y=bx represents decay if the value of b is between 0 and 1. In this range, as x increases, y decreases, indicating a decay or decrease in the value of y. If b is greater than 1, the equation represents growth rather than decay. Similarly, if b is negative, the equation represents a reflection or flip of the graph, but not necessarily decay. Therefore, the correct answer is 0

15. Which of the following is an increasing exponential function whose y-intercept is 50?

Y = 50x - 4

Y = 50(.05)x

Y = 50(1.4)x

Y = (2)x + 50

The correct answer is y = 50(1.4)x. This is an increasing exponential function because the base, 1.4, is greater than 1. As x increases, the value of the function will increase exponentially. The y-intercept is 50, which means that when x is 0, y will be 50.

Explanation

The correct answer is y = 50(1.4)x. This is an increasing exponential function because the base, 1.4, is greater than 1. As x increases, the value of the function will increase exponentially. The y-intercept is 50, which means that when x is 0, y will be 50.