3. Graphing Exponential and Logarithmic Functions
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Chapter 5
3.
Graphing Exponential and Logarithmic Functions
Logarithmic and exponential functions are fundamental mathematical concepts that have various applications in real-world scenarios. The lesson delves into the relationship between their graphs, highlighting their unique characteristics, such as asymptotes. For instance, every logarithmic graph has a vertical asymptote, while every exponential graph has a horizontal one. The site also presents scenarios like calculating atmospheric pressure at different altitudes using natural base exponential functions. Through interactive graphs and exercises, learners can grasp the nuances of these functions and their significance in various fields.
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Lesson Settings & Tools
| | 11 Theory slides |
| | 16 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Graphing Exponential and Logarithmic Functions
Slide of 11
In this lesson, exponential functions with base e and the key features of their graphs will be discussed. Furthermore, logarithmic functions will be introduced and graphed. Finally, the relationship between exponential and logarithmic functions will be explained both algebraically and graphically.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
- Function
- Function notation
- Inverse Function and Finding the Inverse of a Function
- Exponential Function
- Logarithms and Power Property of Logarithms
- Common Logarithm
- Natural Logarithm
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Identifying Change
As soon as Zosia entered the room for her Algebra lesson, her teacher asked a question to the class. 
Help Zosia find the answer by considering the following questions.
- How do the y-values change for one unit increase in x? Is it a constant change?
- Does the table represent a linear relationship?
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Constant Multiplier
The rate of change for linear functions is constant. For each step in the x-direction, the change between y-values is the same. Therefore, the difference between y-values for consecutive x-values is the same. Conversely, the rate of change for exponential functions is not constant. This means that the differences between y-values for consecutive x-values are not the same.
In the table on the left, the difference between y-values for consecutive x-values is 2. This means that the rate of change is constant. Conversely, in the table on the right, the differences between y-values for consecutive x-values are not the same. This means that for this table the rate of change is not constant.
The table on the left represents a linear function because it shows a constant rate of change. Conversely, the table on the right represents a non-linear function. Notice how the y-values, for the table on the right, are doubled for each step in the x-direction. To obtain a y-value, the previous value is multiplied by 2.
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Identifying the Constant Multiplier
Does the table below correspond to an exponential function? If yes, write only the constant multiplier. If not, write only no.
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Natural Base Exponential Functions
Natural base exponential functions are exponential functions whose base, or constant factor, is the number e. f(x) = ae^(rx) Here, a and r are real numbers that are not equal to 0. In considering their positive or negative sign, two observations can be made.
- If both a and r are positive, the function is an exponential growth function.
- If a is positive and r is negative, the resulting function is instead an exponential decay function.
The graphs of f(x)= e^x and g(x) = e^(- x) are shown.
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Graphing Natural Base Functions
A natural base exponential function with both a and r equal to 1 will be considered rather than more complex values for the sake of understanding the basics. y=ae^(rx) [a=1 andr=1]Substitute y=e^x
To graph this function, three steps must be followed.
1
Identify the Initial Value and Plot the y-intercept
expand_more The initial value of an exponential function, y = ab^x, is the number without an exponent, or the value of a. The initial value is also known as the y-intercept (0,a). In this example, the value of a is 1, and thus the y-intercept of the graph of y=e^x is (0,1).
2
Use the Constant Multiplier to Find More Points
expand_more In natural base functions, the constant multiplier — the number with an exponent — is the number e. This means that when x-values increase by 1, the y-values are multiplied by e. With this information, more points can be plotted on the coordinate plane. It is advised to have not less than three points.
3
Draw the Curve
expand_more Lastly, the graph can be drawn by connecting the points with a smooth curve.
It is worth remembering that, in general, the graph of a natural base function is always a smooth curve.
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Continuously Compounded Interest
When interest is compounded infinitely many times, it is said to be continuously compounded. Let A be the balance of an account that is continuously compounded, P the initial amount, r the interest rate, and t the time. These values are connected by the following formula.
A=Pe^(rt)
Keep in mind that, in this formula, the value of r must be written as a decimal and the time t must be in years. Also, the initial amount P is usually called principal.
Proof
Recall the compound interest formula. A = P (1+r/n)^(nt)
In the formula, there is the account's balance A, the initial amount P, the time in years t, and the number of times n the interest is compounded per year.
The formula can be rewritten by using the Power of a Power Property.
A = P (1+r/n)^(nt) ⇕ A = P [ (1+r/n)^n ]^t
If the interest rate is 100 %, or r=1, and the interest is continuously compounded, the formula can be written in terms of P, t, and e. This is because as n goes to infinity, the value of (1+ 1n)^n approaches e.
A = P [ (1 + 1/n)^n ]^t [n→ ∞] A = P e^t
The value of the expression (1+ rn )^n, on the other hand, approaches e^r. Examine the following interactive graph to better grasp how this is possible.
Using this last approximation, the final form of the formula can be obtained.
A = P [ (1 + r/n)^n ]^t [ n→ ∞ ] A = P e^(r t)
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Graphing and Using a Continuously Compounded Interest Graph
Zosia wants to visit family in Argentina, and while there she hopes to climb Mount Aconcagua! This is going to cost a pretty penny so she needs to increase her savings. Zosia knows that banks do not offer continuously compounded interest, but she is daydreaming about opening an account that does use it.
Recently, she made $17 from selling some old stuff at a garage sale. She would deposit this amount into the account. The amount A in the account after t years is calculated by using the following formula. A=17e^t
a Graph the given function.
b Use the formula to calculate the balance of this account after 1 year and 9 months. Suppose that no deposits nor withdrawals are made and round the answer to two decimal places. Check the answer with the graph.
c Use the graph to estimate in how many years the balance of this account would be $ 120. Suppose that no deposits nor withdrawals are made. Round the answer to the nearest integer.
Answer
a
b About $97.83
c About 2 years
Hint
a Start by identifying the initial value and plotting the y-intercept.
b Keep in mind that 1 year and 9 months are 1.75 years.
c Identify the point on the curve whose second coordinate is 120.
Solution
a The given function is a natural base exponential function whose initial value is 17.
A= 17e^t Therefore, the y-intercept of the graph occurs at (0, 17). Since time cannot be negative and the balance of the account will always be non-negative, only the first quadrant of the coordinate plane will be considered.
The constant multiplier is e. This means that when x-values increase by 1, the y-values are multiplied by e.
Finally, the curve can be graphed by connecting these points.
b First, 1 year and 9 months must be expressed in years. Recall that 1 year has 12 months. This means that 9 months are 912, or 0.75, years.
A=17e^t
Substitute
t= 1.75
A=17e^(1.75)
UseCalc
Use a calculator
A=97.828245...
RoundDec
Round to 2 decimal place(s)
A≈ 97.83
The second coordinate of this point is a bit less than 100. Therefore, it is reasonable to have a balance of $97.83 after 1 year and 9 months.
c This time the graph will be used to estimate after how many years the balance would be $120. To do so, the point with the second coordinate 120 must be spotted on the graph of the function, and its first coordinate needs to be identified.
The first coordinate of the point is almost equal to 2. Therefore, the balance would be $120 after about 2 years. Zosia is realizing that it might take her quite some time to meet her financial goal, but she is making a valiant effort and certainly on her way!
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Using a Natural Base Exponential Function to Model the Atmospheric Pressure at Different Altitudes
Zosia finally saved enough money to visit family in Argentina and set out to climb Mount Aconcagua while there! She is as excited as ever.
Here, her knowledge about natural base exponential functions will be extremely useful to her pursuit of reaching the peak. Zosia knows that the atmospheric pressure decreases as the altitude increases, the pressure at sea level is about 1013 hecto Pascals (hPa), and the pressure P(x) at an altitude of x meters is given by the following formula. P(x)=1013e^(- 0.000128x) To avoid unforeseen problems, Zosia realizes that she can share some responsibilities if she recruits a team to solve such issues. Give her a hand and become a valuable member of her team by answering the following questions. If unable to, the team could get altitude sickness!
a The altitude of the mountain's peak, called the summit, is 8849 meters. Calculate the atmospheric pressure at the summit. Round the answer to the nearest integer.
b Draw a graph of the given exponential function.
c Estimate the altitude where the atmospheric pressure is 700 hecto Pascals. At this altitude, Zosia will rest for a night or two. Round the answer to the nearest thousand meters.
Answer
a About 326 hPa
b
c About 3000 m
Hint
a Substitute 8849 for x in the given formula.
b Make a table of values.
c Use the graph from Part B.
Solution
a To calculate the atmospheric pressure at an altitude of 8849 meters, this number will be substituted for x into the given function.
P(x)=1013e^(- 0.000128x)
Substitute
x= 8849
P( 8849)=1013e^(- 0.000128( 8849))
▼
Evaluate right-hand side
MultNegPos
(- a)b = - ab
P(8849)=1013e^(- 1.132672)
UseCalc
Use a calculator
P(8849)=326.359490...
RoundInt
Round to nearest integer
P(8849)≈ 326
b To graph the given function, a table of values will be made first. Only non-negative values will be considered since it is a mountain climb under consideration.
| x | 1013e^(- 0.000128x) | y=1013e^(- 0.000128x) |
|---|---|---|
| 0 | 1013e^(- 0.000128( 0)) | 1013 |
| 1000 | 1013e^(- 0.000128( 1000)) | ≈ 891 |
| 2000 | 1013e^(- 0.000128( 2000)) | ≈ 784 |
| 3000 | 1013e^(- 0.000128( 3000)) | ≈ 690 |
| 4000 | 1013e^(- 0.000128( 4000)) | ≈ 607 |
| 5000 | 1013e^(- 0.000128( 5000)) | ≈ 534 |
| 6000 | 1013e^(- 0.000128( 6000)) | ≈ 470 |
| 7000 | 1013e^(- 0.000128( 7000)) | ≈ 414 |
| 8000 | 1013e^(- 0.000128( 8000)) | ≈ 364 |
| 9000 | 1013e^(- 0.000128( 9000)) | ≈ 320 |
| 10000 | 1013e^(- 0.000128( 10000)) | ≈ 282 |
The obtained points will now be plotted on a coordinate plane and connected with a smooth curve. Since only non-negative values are considered, the graph just needs to be drawn in the first quadrant.
c To estimate the altitude where the atmospheric pressure is 700 hecto Pascals, the graph from Part B will be used. The point with the second coordinate 700 will be spotted and its x-coordinate identified.
Tracing a pen or pencil vertically down from the point of the graph that is even, horizontally, with y-value of 700 leads to the x-value of 3. This means that the altitude with an atmospheric pressure of 700 hecto Pascals is about 3000 meters. Zosia feels comfortable knowing she will be able to rest for a few nights at this level.
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Logarithmic Functions
Logarithmic functions are functions that involve logarithms.
f(x) = log_b x, b>0andb≠ 1
The function f(x)=log_bx is the parent function of logarithmic functions. Since logarithms are defined for positive numbers, the domain of the function is x>0 and its range is all real numbers. If b is less than 1, the graph of the function is decreasing over its entire domain. Conversely, if b is greater than 1, the graph is increasing over its entire domain.
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Inverse Properties of Logarithms
A logarithm and a power with the same base undo
each other.
log_b b^x =x and b^(log_b x)=x
In particular, the above equations also hold true for common and natural logarithms.
rcr log 10^x=x & and & 10^(log x)=x [0.8em] ln e^x=x & and& e^(ln x)=x
Proof
The general equations will be proved one at a time.
log_b b^x =x
This identity can be proved by using the Power Property of Logarithms and the definition of a logarithm.log_b b^x
log_b(a^m)= m* log_b(a)
x log_b b
log_b(b) = 1
x(1)
IdPropMult
Identity Property of Multiplication
x ✓
b^(log_b x)=x
Let log_b x=a. Therefore, by the definition of a logarithm, b^a=x. log_b x=a ⇔ b^a=x This will be used to prove the identity. Therefore, b to the power of log_bx is equal to x.Consider the following exponential and logarithmic functions. f(x)= b^x and g(x)= log_b x By using the Inverse Property of Logarithms, the composition of these functions results in the identity function. c|c f( g(x) )= b^(log_b x) & g( f(x)) = log_b b^x ⇕ & ⇕ f(g(x) ) = x & g(f(x)) = x This means that exponential and logarithmic functions are inverse functions. Therefore, the graphs of these functions are each other's reflection across the line y=x.
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Modeling Battery Charges Using Logarithmic Functions
Zosia is getting everything ready for the multi-day hike in Mount Aconcagua. Because her extra battery pack is quite limited, she wants her smartphone and camera to have their batteries fully charged.
Answer
Phone's Battery:
Camera's Battery:
Hint
To graph p(x)=ln x, first graph its inverse function g(x)=e^x and then reflect the curve across the line y=x. To graph c(x)=2log (x-0.5), make a table of values. Consider its domain first.
Solution
The logarithmic functions will be graphed one at a time.
Graphing p(x)=ln x
It is known that logarithms and exponents are inverse operations. Now, consider the natural base exponential function g(x)=e^x. By the Inverse Properties of Logarithms, it can be shown that the compositions of p(x) and g(x) result in the identity function. c|c p(g(x))=ln e^x & g(p(x))=e^(ln x) ⇕ & ⇕ p(g(x))=x & g(p(x))=x Therefore, p(x) and g(x) are inverse functions. This means that the graph of p(x) is a reflection across the line y=x of the graph of g(x). Since g(x) is a natural base exponential function, the graph can be drawn by using its initial value and constant multiplier.
Graphing c(x)=2log (x-0.5)
To graph this function, a table of values will be used. Before that, however, the domain of c(x) should be determined. To do that, recall that logarithms are defined only for positive numbers. Therefore, whatever the expression is in the logarithm, it must be positive. c(x)=2log ( x-0.5) [0.8em] x-0.5 >0 ⇔ x>0.5 The domain of the function is the set of all real numbers greater than 0.5. To make the table, only values in the domain will be considered.
| x | 2log (x-0.5) | c(x)=2log (x-0.5) |
|---|---|---|
| 1 | 2log ( 1-0.5) | ≈ - 0.6 |
| 2 | 2log ( 2-0.5) | ≈ 0.35 |
| 3 | 2log ( 3-0.5) | ≈ 0.8 |
| 4 | 2log ( 4-0.5) | ≈ 1.1 |
| 5 | 2log ( 5-0.5) | ≈ 1.3 |
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Finding Inverse Functions of Logarithmic and Exponential Functions
The expedition was a complete success! What is more, some members of the hiking team felt so inspired by Zosia’s math skills in helping them reach the summit, that they decided to practice some logarithmic functions. They believe this knowledge will also help them in their next expedition to the Amazon Rainforest.
Touched by their passion to learn, Zosia found a few interesting exercises about exponential and logarithmic functions in her online textbook to share with the team.
a Find the inverse function of f(x)=2log (x-1).
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b Find the inverse function of g(x)= 15e^(3x).
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Hint
a The inverse of f(x) is a common base exponential function.
b The inverse of g(x) is a natural base logarithmic function.
Solution
a To find the inverse of the function, the first step is replacing f(x) with y.
10^(12x)=y-1
AddEqn
LHS+1=RHS+1
10^(12x)+1=y
RearrangeEqn
Rearrange equation
y=10^(12x)+1
Substitute
y= f^(- 1)(x)
f^(- 1)(x)=10^(12x)+1
| x | 2log (x-1) | f(x) |
|---|---|---|
| 1.5 | 2log ( 1.5-1) | ≈ - 0.6 |
| 5 | 2log ( 5-1) | ≈ 1.2 |
| 10 | 2log ( 10-1) | ≈ 1.9 |
| 12 | 2log ( 12-1) | ≈ 2.1 |
| 18 | 2log ( 18-1) | ≈ 2.5 |
Next, the table that corresponds to f^(- 1)(x) will be constructed.
| x | 10^(12x)+1 | f^(- 1)(x) |
|---|---|---|
| - 2 | 10^(12( - 2))+1 | 1.1 |
| - 1 | 10^(12( - 1))+1 | ≈ 1.3 |
| 0 | 10^(12( 0))+1 | 2 |
| 1 | 10^(12( 1))+1 | ≈ 4.2 |
| 2 | 10^(12( 2))+1 | 11 |
Now the points can be plotted on a coordinate plane and connected with smooth curves. Also, the line y=x will be graphed.
The graphs are each other's reflection across the line y=x. Therefore, f^(-1)(x)=10^(12x)+1 is the inverse function of f(x)=2log (x-1).
b As in Part A, to find the inverse of the function the first step is replacing g(x) with y.
ln 5x=3y
DivEqn
.LHS /3.=.RHS /3.
ln 5x/3=y
MoveNumRight
a/b=1/b* a
1/3ln 5x=y
RearrangeEqn
Rearrange equation
y=1/3ln 5x
Substitute
y= g^(- 1)(x)
g^(- 1)(x)=1/3ln 5x
| x | 1/5e^(3x) | g(x) |
|---|---|---|
| - 1 | 1/5e^(3( - 1)) | ≈ 0.01 |
| - 0.5 | 1/5e^(3( - 0.5)) | ≈ 0.04 |
| 0 | 1/5e^(3( 0)) | 0.2 |
| 0.5 | 1/5e^(3( 0.5)) | ≈ 0.9 |
| 1 | 1/5e^(3( 1)) | ≈ 4 |
Next, the table that corresponds to g^(- 1)(x) will be constructed.
| x | 1/3ln 5x | g^(- 1)(x) |
|---|---|---|
| 0.1 | 1/3ln 5( 0.5) | ≈ - 0.2 |
| 1 | 1/3ln 5( 1) | ≈ 0.5 |
| 2 | 1/3ln 5( 2) | ≈ 0.8 |
| 3 | 1/3ln 5( 3) | ≈ 0.9 |
| 4 | 1/3ln 5( 4) | ≈ 1 |
The obtained points can be plotted on a coordinate plane and connected with smooth curves. Also, the line y=x will be graphed.
It can be seen that the graphs are each other's reflection across the line y=x. Therefore, g^(-1)(x)= 13ln 5x is the inverse function of g(x)= 15e^(3x).
Checking Our Answer
a To verify that the obtained function is actually the inverse of the given function, instead of graphing it can be checked that both composite functions f(f^(- 1)(x)) and f^(- 1)(f(x)) simplify to x. Start by calculating f(f^(- 1)(x)).
f(f^(- 1)(x))=2log (f^(- 1)(x)-1)
Substitute
f^(- 1)(x)= 10^(12x)+1
f(f^(- 1)(x))=2log ( 10^(12x)+1-1)
▼
Simplify right-hand side
SubTerm
Subtract term
f(f^(- 1)(x))=2log (10^(12x))
log(10^m)=m
f(f^(- 1)(x))=2(1/2x)
AssociativePropMult
Associative Property of Multiplication
f(f^(- 1)(x))=2(1/2)x
DenomMultFracToNumber
2 * a/2= a
f(f^(- 1)(x))=1x
IdPropMult
Identity Property of Multiplication
f(f^(- 1)(x))=x ✓
f^(- 1)(f(x))=10^(12f(x))+1
Substitute
f(x)= 2log(x-1)
f^(- 1)(f(x))=10^(12( 2log(x-1)))+1
▼
Simplify right-hand side
AssociativePropMult
Associative Property of Multiplication
f^(- 1)(f(x))=10^(12(2)log(x-1))+1
FracMultDenomToNumber
a/2* 2 = a
f^(- 1)(f(x))=10^(1log(x-1))+1
IdPropMult
Identity Property of Multiplication
f^(- 1)(f(x))=10^(log(x-1))+1
10^(log(m))=m
f^(- 1)(f(x))=x-1+1
AddTerms
Add terms
f^(- 1)(f(x))=x ✓
b As in the previous part, to check whether g^(- 1)(x)= 13ln 5x is the inverse function of g(x)= 15e^(3x), the compositions g(g^(- 1)(x)) and g^(- 1)(g(x)) must both simplify to x.
| Definition of the First Function | Substitute the Second Function | Simplify | |
|---|---|---|---|
| g(g^(- 1)(x)) | 1/5e^(3g^(-1)(x)) | 1/5e^(3( 13ln 5x)) | x ✓ |
| g^(- 1)(g(x)) | 1/3ln 5g(x) | 1/3ln 5( 15e^(3x)) | x ✓ |
Since both compositions simplify to the identity function, g^(- 1)(x)= 13ln 5x is the inverse function of g(x)= 15e^(3x).
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Horizontal and Vertical Asymptotes
In this lesson, exponential functions, logarithmic functions, and the relationship between their graphs have been discussed. One more characteristic of these graphs is that they have asymptotes.
Exponential Graphs
Every exponential graph has a horizontal asymptote. In the applet below, both graphs have the horizontal line y=0, which is the x-axis, as an asymptote.
Logarithmic Graphs
Every logarithmic graph has a vertical asymptote. In the applet below, both graphs have the vertical line x=0, which is the y-axis, as an asymptote.
Graphing Exponential and Logarithmic Functions
Exercise 1.1
Suppose that the ratio between consecutive y-values in a table is always the same. In such a case, the function represented by the table is called an exponential function. Additionally, its constant multiplier c is the quotient between two consecutive y-values.
Determine whether the following tables represent an exponential function. If yes, write the constant multiplier. Otherwise, just write no.
| x | y |
|---|---|
| 1 | 2 |
| 2 | 3 |
| 3 | 4.5 |
| 4 | 6.75 |
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| x | y |
|---|---|
| 1 | 5 |
| 2 | 10 |
| 3 | 15 |
| 4 | 20 |
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Let's find the ratio between consecutive y-values.
We can see that the ratio between consecutive y-values is constant and equal to 1.5. Therefore, the table does represent an exponential function with a constant multiplier of 1.5
Let's find the ratio between consecutive y-values.
This time, the ratios are not constant. Therefore, the table does not represent an exponential function.
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An investment account for a house earns interest that is compounded continuously. The balance B of the account, in dollars, after t years is modeled by the following function.
B=3224e^(0.05t)
What is the balance after 10 years? Write the answer to the nearest dollar.
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We want to find the balance of the account after 10 years. To do so, we will substitute 10 for t in the given continuously compounded interest formula. Let's do it!
After 10 years, the balance of the account will be $5315 when rounded to the nearest dollar.
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An investment account for a house earns annual interest compounded continuously. The balance B of the account in dollars after t years is modeled by the following function. B=3224e^(0.05t) Which of the following graphs is the graph of this function?
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The given formula represents a continuously compounded interest. In this formula, the interest rate and the principal are both positive.
| B=3224e^(0.05t) | |
|---|---|
| Initial Value | Interest Rate |
| 3224 | 0.05 |
This means that the graph of this function is increasing in its entire domain. Therefore, we can disregard options A and C, since they represent decreasing functions.
Next, let's find the y-intercept of the graph. To do so, we will substitute 0 for t in our formula. Let's do it!
The y-intercept of the graph occurs at (0,3224), so the graph intersects the y-axis above the x-axis. This characteristic matches the graph in choice B.
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A savings account earns compound interest. The balance of the account B after t years is represented by the following graph.
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To find the balance of the account after 3 years, we will identify the point on the curve whose x-coordinate is 3. The balance after 3 years is the y-coordinate of this point.
The exact value of the y-coordinate cannot be determined by only looking at the graph. However, we can see that, to the nearest thousand, the balance of the account after 3 years will be about $6000.
To find after how many years the balance of the account will be $ 8000, we will identify the point on the curve whose y-coordinate is 8000. Then, we will determine its x-coordinate.
The exact value of the x-coordinate cannot be determined by only using the graph. However, we can see that, to the nearest integer, the x-coordinate is 6. Therefore, the balance of the account will be $8000 after about 6 years.
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Atmospheric pressure decreases as altitude increases. The pressure at sea level is about 1013 hecto Pascals (hPa), and the pressure P(x) at an altitude of x meters is given by the following formula.
P(x)=1013e^(- 0.000128x)
Find the atmospheric pressure in the city of La Paz, Bolivia, which is located at an altitude of 3625 meters above sea level. Round the answer to the nearest integer.
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We want to find the atmospheric pressure in La Paz, which has an altitude of 3625 meters above sea level. To do so, we will substitute 3625 for x in the given natural base exponential function. Let's do it!
Therefore, the atmospheric pressure in La Paz is about 637 hecto Pascals.
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Atmospheric pressure decreases as altitude increases. The pressure at sea level is about 1013 hecto Pascals (hPa), and the pressure P(x) at an altitude of x meters is given by the following formula. P(x)=1013e^(- 0.000128x) Which of the following graphs can be the graph of this function?
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The given function is a natural base exponential function. In this formula, the coefficient of the variable in the exponent is negative.
| P(x)=1013e^(- 0.000128x) | |
|---|---|
| Initial Value | Coefficient of x |
| 1013 | - 0.000128 |
This means that the graph of this function is decreasing in its entire domain. Therefore, we can disregard options B and D, since they represent increasing functions.
Next, let's find the y-intercept of the graph. To do so, we will substitute 0 for x in our formula. Let's do it!
The y-intercept of the graph occurs at (0,1013), so the graph intersects the y-axis above the x-axis. This characteristic matches the graph in choice C.
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The atmospheric pressure P(x), in hecto Pascals, at x meters above sea level is represented by the following graph.
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Let's use the graph to find the atmospheric pressure at an altitude of 12 000 meters above sea level. To do so, we will identify the point on the curve whose first coordinate is 12 000.
We cannot state the exact value for the second coordinate of this point. However, we can see that the second coordinate of this point, rounded to the nearest hundred, is 200. This means that the atmospheric pressure at an altitude of 12 000 meters is about 200 hecto Pascals.
This time we need to find the altitude when the atmospheric pressure is 600 hecto Pascals. To do so, we will identify the point on the curve whose second coordinate is 600.
We cannot state the exact value for the first coordinate of this point. However, we can see that the first coordinate of this point, rounded to the nearest thousand, is 4000. This means that the altitude when the atmospheric pressure is 600 hecto Pascals is about 4000 meters.
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Consider the given logarithmic function. f(x)=log (x-1) Which of the following is the graph of the function?
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To determine the graph of the given logarithmic function, we will start by finding its domain. To do so, recall that the argument of a logarithm must be positive. log( x-1_(> 0) ) ⇒ x>1 At this point we can say that the domain of the function is the set of real numbers greater than 1. Therefore, the graph of the function must be located to the right of the vertical line x=1, which is a vertical asymptote. With this information, we can disregard graphs C and D.
Now, the possible choices are only A and B. Notice that the graph shown in A is an increasing curve, while the graph shown in B is a decreasing curve. Therefore, to determine which option is the correct choice, we will evaluate the function at two values of x. For convenience, we can choose x=2 and x=11. Let's start by evaluating the function at x=2.
We found that f(2)=0, so the x-intercept of the graph occurs at x=2. Now let's calculate the value of the function for x=11. If the result is positive, the graph shown in A is the correct choice because the values of this curve to the right of the x-intercept are positive. Otherwise, the graph shown in B is the correct choice, as the values of this curve to the right of the x-intercept are negative.
We found that f(11) is positive, so the graph that represents the given function is the graph shown in A.
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Consider the given logarithmic function. g(x)=ln (3-x) Which of the following is the graph of the function?
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To determine which of the graphs is the graph of the logarithmic function, we will draw its graph and compare it with the given choices. To do so, we will start by finding the domain. Recall that the argument of a logarithm must be positive. Function g(x) = ln( 3-x) [0.7em] Domain 3-x>0 ⇔ x<3 The domain of the function is the set of real numbers less than 3. Let's use this to make a table of values.
| x | ln (3-x) | g(x) |
|---|---|---|
| 2.5 | ln (3- 2.5) | ≈ - 0.69 |
| 2 | ln (3- 2) | 0 |
| 1 | ln (3- 1) | ≈ 0.69 |
| 0 | ln (3- 0) | ≈ 1.1 |
| - 1 | ln (3-( - 1)) | ≈ 1.39 |
We can now plot the points obtained in the table and connect them with a smooth curve. Remember that the domain is the set of all real numbers less than 3.
This graph matches option A.
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Consider the logarithmic function.
f(x)=ln (2x+1)
Find f^(- 1)(x), the inverse function of f(x).
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To find the inverse of the logarithmic function, we will start by replacing f(x) with y. y=ln (2x+1) Next, we will switch the variables x and y. This will give us the inverse of the function. Function y=ln (2 x+1) [0.3em] ⇓ [0.3em] Switch Variables x=ln (2 y+1) Now we will isolate the y-variable. To do so, we can use the definition of a logarithm. Recall that a natural logarithm is a logarithm with base e. x=ln (2y+1) ⇔ e^x=2y+1 We can use inverse operations to finally isolate y.
Therefore, the inverse of the given function is f^(- 1)(x)= e^x-12.
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Consider the given exponential function.
g(x)=1/2 (10^(13x))
Find g^(- 1)(x), the inverse function of g(x).
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To find the inverse of the logarithmic function, we will start by replacing g(x) with y. y=1/2 (10^(13x)) Next, we will switch the variables x and y. Function y=1/2 (10^(13 x)) [0.6em] ⇓ [0.4em] Switch Variables x=1/2 (10^(13 y)) Now we will isolate the y-variable. To do so, we can first multiply both sides of the equation by 2.
We can use the definition of a logarithm. Recall that a logarithm with base 10 is a common logarithm. 2x=10^(13y) ⇔ log 2x=1/3y Finally, we can multiply both sides of the equation by 3 and simplify by using the Power Property of Logarithms.
Therefore, the inverse of the given function is g^(- 1)(x)=log 8x^3.
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Consider the logarithmic function.
y=log_2 (4x-32)
Find the domain of this function. Write the answer as an inequality.
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To find the domain of the logarithmic function, we have to keep in mind that the argument of a logarithm must be positive.
Therefore, the domain of the function is the set of all real numbers greater than 8.
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Graphing Exponential and Logarithmic Functions
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