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Chapter 6
Exponential and Logarithmic
Functions
6.1 Introduction to Exponential and Logarithmic Functions
Of all of the functions we study in this text, exponential and logarithmic functions are possibly the ones which impact everyday life the most.^1 This section will introduce us to these functions while the rest of the chapter will more thoroughly explore their properties. Up to this point, we have dealt with functions which involve terms like x^2 or x^2 /^3 , in other words, terms of the form xp where the base of the term, x, varies but the exponent of each term, p, remains constant. In this chapter, we study functions of the form f (x) = bx^ where the base b is a constant and the exponent x is the variable. We start our exploration of these functions with f (x) = 2x. (Apparently this is a tradition. Every College Algebra book we have ever read starts with f (x) = 2x.) We make a table of values, plot the points and connect them in a pleasing fashion.
x f (x) (x, f (x)) − 3 2 −^3 = (^18)
− 2 2 −^2 = 14
− 1 2 −^1
x
y
− 3 − 2 − 1 1 2 3
1
2
3
4
5
6
7
8
y = f (x) = 2x
A few remarks about the graph of f (x) = 2x^ which we have constructed are in order. As x → −∞
(^1) Take a class in Differential Equations and you’ll see why.
418 Exponential and Logarithmic Functions
and attains values like x = −100 or x = −1000, the function f (x) = 2x^ takes on values like f (−100) = 2−^100 = 21001 or f (−1000) = 2−^1000 = 210001. In other words, as x → −∞,
2 x^ ≈
very big (+) ≈ very small (+)
So as x → −∞, 2x^ → 0 +. This is represented graphically using the x-axis (the line y = 0) as a horizontal asymptote. On the flip side, as x → ∞, we find f (100) = 2^100 , f (1000) = 2^1000 , and so on, thus 2x^ → ∞. As a result, our graph suggests the range of f is (0, ∞). The graph of f passes the Horizontal Line Test which means f is one-to-one and hence invertible. We also note that when we ‘connected the dots in a pleasing fashion’, we have made the implicit assumption that f (x) = 2x is continuous^2 and has a domain of all real numbers. In particular, we have suggested that things like 2
√ (^3) exist as real numbers. We should take a moment to discuss what something like 2 √ (^3) might
mean, and refer the interested reader to a solid course in Calculus for a more rigorous explanation. The number
3 = 1. 73205... is an irrational number^3 and as such, its decimal representation neither repeats nor terminates. We can, however, approximate
3 by terminating decimals, and it stands to reason^4 we can use these to approximate 2
√ (^3). For example, if we approximate √ 3
by 1.73, we can approximate 2
√ 3 ≈ 21.^73 = 2 (^173100) = 100
- It is not, by any means, a pleasant number, but it is at least a number that we understand in terms of powers and roots. It also stands to reason that better and better approximations of
3 yield better and better approximations of 2
√ (^3) , so the value of 2 √ (^3) should be the result of this sequence of approximations. 5
Suppose we wish to study the family of functions f (x) = bx. Which bases b make sense to study? We find that we run into difficulty if b < 0. For example, if b = −2, then the function f (x) = (−2)x has trouble, for instance, at x = 12 since (−2)^1 /^2 =
−2 is not a real number. In general, if x is any rational number with an even denominator, then (−2)x^ is not defined, so we must restrict our attention to bases b ≥ 0. What about b = 0? The function f (x) = 0x^ is undefined for x ≤ 0 because we cannot divide by 0 and 0^0 is an indeterminant form. For x > 0, 0x^ = 0 so the function f (x) = 0x^ is the same as the function f (x) = 0, x > 0. We know everything we can possibly know about this function, so we exclude it from our investigations. The only other base we exclude is b = 1, since the function f (x) = 1x^ = 1 is, once again, a function we have already studied. We are now ready for our definition of exponential functions.
Definition 6.1. A function of the form f (x) = bx^ where b is a fixed real number, b > 0, b 6 = 1 is called a base b exponential function.
We leave it to the reader to verify^6 that if b > 1, then the exponential function f (x) = bx^ will share the same basic shape and characteristics as f (x) = 2x. What if 0 < b < 1? Consider g(x) =
2
)x . We could certainly build a table of values and connect the points, or we could take a step back and
(^2) Recall that this means there are no holes or other kinds of breaks in the graph. (^3) You can actually prove this by considering the polynomial p(x) = x (^2) − 3 and showing it has no rational zeros by applying Theorem 3.9. (^4) This is where Calculus and continuity come into play. (^5) Want more information? Look up “convergent sequences” on the Internet. (^6) Meaning, graph some more examples on your own.
420 Exponential and Logarithmic Functions
Of all of the bases for exponential functions, two occur the most often in scientific circles. The first, base 10, is often called the common base. The second base is an irrational number, e ≈ 2 .718, called the natural base. We will more formally discuss the origins of this number in Section 6.5. For now, it is enough to know that since e > 1, f (x) = ex^ is an increasing exponential function. The following examples give us an idea how these functions are used in the wild.
Example 6.1.1. The value of a car can be modeled by V (x) = 25
5
)x , where x ≥ 0 is age of the car in years and V (x) is the value in thousands of dollars.
- Find and interpret V (0).
- Sketch the graph of y = V (x) using transformations.
- Find and interpret the horizontal asymptote of the graph you found in 2.
Solution.
- To find V (0), we replace x with 0 to obtain V (0) = 25
5
= 25. Since x represents the age of the car in years, x = 0 corresponds to the car being brand new. Since V (x) is measured in thousands of dollars, V (0) = 25 corresponds to a value of $25,000. Putting it all together, we interpret V (0) = 25 to mean the purchase price of the car was $25,000.
- To graph y = 25
5
)x , we start with the basic exponential function f (x) =
5
)x
. Since the base b = 45 is between 0 and 1, the graph of y = f (x) is decreasing. We plot the y-intercept (0, 1) and two other points,
and
, and label the horizontal asymptote y = 0. To obtain V (x) = 25
5
)x , x ≥ 0, we multiply the output from f by 25, in other words, V (x) = 25f (x). In accordance with Theorem 1.5, this results in a vertical stretch by a factor of 25. We multiply all of the y values in the graph by 25 (including the y value of the horizontal asymptote) and obtain the points
, (0, 25) and (1, 20). The horizontal asymptote remains y = 0. Finally, we restrict the domain to [0, ∞) to fit with the applied domain given to us. We have the result below.
(0, 1)
H.A. y = 0
x
y
− 3 − 2 − 1 1 2 3
2
y = f (x) = ( (^4) 5
)x
vertical scale by a factor of 25 −−−−−−−−−−−−−−−−−−−−−→ multiply each y-coordinate by 25
(0, 25)
H.A. y = 0
x
y
1 2 3 4 5 6
5
10
15
20
30
y = V (x) = 25f (x), x ≥ 0
- We see from the graph of V that its horizontal asymptote is y = 0. (We leave it to reader to verify this analytically by thinking about what happens as we take larger and larger powers of 45 .) This means as the car gets older, its value diminishes to 0.
6.1 Introduction to Exponential and Logarithmic Functions 421
The function in the previous example is often called a ‘decay curve’. Increasing exponential func- tions are used to model ‘growth curves’ and we shall see several different examples of those in Section 6.5. For now, we present another common decay curve which will serve as the basis for further study of exponential functions. Although it may look more complicated than the previ- ous example, it is actually just a basic exponential function which has been modified by a few transformations from Section 1.7.
Example 6.1.2. According to Newton’s Law of Cooling^8 the temperature of coffee T (in degrees Fahrenheit) t minutes after it is served can be modeled by T (t) = 70 + 90e−^0.^1 t.
- Find and interpret T (0).
- Sketch the graph of y = T (t) using transformations.
- Find and interpret the horizontal asymptote of the graph.
Solution.
- To find T (0), we replace every occurrence of the independent variable t with 0 to obtain T (0) = 70 + 90e−^0 .1(0)^ = 160. This means that the coffee was served at 160◦F.
- To graph y = T (t) using transformations, we start with the basic function, f (t) = et. As we have already remarked, e ≈ 2. 718 > 1 so the graph of f is an increasing exponential with y-intercept (0, 1) and horizontal asymptote y = 0. The points
− 1 , e−^1
≈ (− 1 , 0 .37) and (1, e) ≈ (1, 2 .72) are also on the graph. Since the formula T (t) looks rather complicated, we rewrite T (t) in the form presented in Theorem 1.7 and use that result to track the changes to our three points and the horizontal asymptote. We have
T (t) = 70 + 90e−^0.^1 t^ = 90e−^0.^1 t^ + 70 = 90f (− 0. 1 t) + 70
Multiplication of the input to f , t, by − 0 .1 results in a horizontal expansion by a factor of 10 as well as a reflection about the y-axis. We divide each of the x values of our points by − 0. 1 (which amounts to multiplying them by −10) to obtain
10 , e−^1
, (0, 1), and (− 10 , e). Since none of these changes affected the y values, the horizontal asymptote remains y = 0. Next, we see that the output from f is being multiplied by 90. This results in a vertical stretch by a factor of 90. We multiply the y-coordinates by 90 to obtain
10 , 90 e−^1
, (0, 90), and (− 10 , 90 e). We also multiply the y value of the horizontal asymptote y = 0 by 90, and it remains y = 0. Finally, we add 70 to all of the y-coordinates, which shifts the graph upwards to obtain
10 , 90 e−^1 + 70
≈ (10, 103 .11), (0, 160), and (− 10 , 90 e + 70) ≈ (− 10 , 314 .64). Adding 70 to the horizontal asymptote shifts it upwards as well to y = 70. We connect these three points using the same shape in the same direction as in the graph of f and, last but not least, we restrict the domain to match the applied domain [0, ∞). The result is below. (^8) We will discuss this in greater detail in Section 6.5.
6.1 Introduction to Exponential and Logarithmic Functions 423
y = bx, b > 1 y = logb(x), b > 1
y = bx, 0 < b < 1 y = logb(x), 0 < b < 1
On a procedural level, logs undo the exponentials. Consider the function f (x) = 2x. When we evaluate f (3) = 2^3 = 8, the input 3 becomes the exponent on the base 2 to produce the real number 8. The function f −^1 (x) = log 2 (x) then takes the number 8 as its input and returns the exponent 3 as its output. In symbols, log 2 (8) = 3. More generally, log 2 (x) is the exponent you put on 2 to get x. Thus, log 2 (16) = 4, because 2^4 = 16. The following theorem summarizes the basic properties of logarithmic functions, all of which come from the fact that they are inverses of exponential functions.
Theorem 6.2. Properties of Logarithmic Functions: Suppose f (x) = logb(x).
- The domain of f is (0, ∞) and the range of f is (−∞, ∞).
- (1, 0) is on the graph of f and x = 0 is a vertical asymptote of the graph of f.
- f is one-to-one, continuous and smooth
- ba^ = c if and only if logb(c) = a. That is, logb(c) is the exponent you put on b to obtain c.
- logb (bx) = x for all x and blogb(x)^ = x for all x > 0
- If b > 1:
- f is always increasing
- As x → 0 +, f (x) → −∞
- As x → ∞, f (x) → ∞
- The graph of f resembles:
y = logb(x), b > 1
- If 0 < b < 1:
- f is always decreasing
- As x → 0 +, f (x) → ∞
- As x → ∞, f (x) → −∞
- The graph of f resembles:
y = logb(x), 0 < b < 1
424 Exponential and Logarithmic Functions
As we have mentioned, Theorem 6.2 is a consequence of Theorems 5.2 and 5.3. However, it is worth the reader’s time to understand Theorem 6.2 from an exponential perspective. For instance, we know that the domain of g(x) = log 2 (x) is (0, ∞). Why? Because the range of f (x) = 2x^ is (0, ∞). In a way, this says everything, but at the same time, it doesn’t. For example, if we try to find log 2 (−1), we are trying to find the exponent we put on 2 to give us −1. In other words, we are looking for x that satisfies 2x^ = −1. There is no such real number, since all powers of 2 are positive. While what we have said is exactly the same thing as saying ‘the domain of g(x) = log 2 (x) is (0, ∞) because the range of f (x) = 2x^ is (0, ∞)’, we feel it is in a student’s best interest to understand the statements in Theorem 6.2 at this level instead of just merely memorizing the facts.
Example 6.1.3. Simplify the following.
- log 3 (81) (^) 2. log 2
- log√ 5 (25) (^) 4. ln
√ (^3) e 2
- log(0.001) 6. 2log^2 (8)^ 7. 117−^ log^117 (6)
Solution.
- The number log 3 (81) is the exponent we put on 3 to get 81. As such, we want to write 81 as a power of 3. We find 81 = 3^4 , so that log 3 (81) = 4.
- To find log 2
8
, we need rewrite 18 as a power of 2. We find 18 = 213 = 2−^3 , so log 2
8
- To determine log√ 5 (25), we need to express 25 as a power of
- We know 25 = 5^2 , and 5 =
, so we have 25 =
. We get log√ 5 (25) = 4.
- First, recall that the notation ln
e^2
means loge
e^2
, so we are looking for the exponent to put on e to obtain 3
e^2. Rewriting 3
e^2 = e^2 /^3 , we find ln
e^2
= ln
e^2 /^3
- Rewriting log(0.001) as log 10 (0.001), we see that we need to write 0.001 as a power of 10. We have 0.001 = 10001 = 1013 = 10−^3. Hence, log(0.001) = log
10 −^3
- We can use Theorem 6.2 directly to simplify 2log^2 (8)^ = 8. We can also understand this problem by first finding log 2 (8). By definition, log 2 (8) is the exponent we put on 2 to get 8. Since 8 = 2^3 , we have log 2 (8) = 3. We now substitute to find 2log^2 (8)^ = 2^3 = 8.
- From Theorem 6.2, we know 117log^117 (6)^ = 6, but we cannot directly apply this formula to the expression 117−^ log^117 (6). (Can you see why?) At this point, we use a property of exponents followed by Theorem 6.2 to get^9
117 −^ log^117 (6)^ =
117 log^117 (6)^
(^9) It is worth a moment of your time to think your way through why 117log 117 (6) (^) = 6. By definition, log 117 (6) is the exponent we put on 117 to get 6. What are we doing with this exponent? We are putting it on 117. By definition we get 6. In other words, the exponential function f (x) = 117x^ undoes the logarithmic function g(x) = log 117 (x).
426 Exponential and Logarithmic Functions
y = f (x) = 2 log(3 − x) − 1 y = g(x) = ln
x x − 1
While logarithms have some interesting applications of their own which you’ll explore in the exer- cises, their primary use to us will be to undo exponential functions. (This is, after all, how they were defined.) Our last example solidifies this and reviews all of the material in the section.
Example 6.1.5. Let f (x) = 2x−^1 − 3.
- Graph f using transformations and state the domain and range of f.
- Explain why f is invertible and find a formula for f −^1 (x).
- Graph f −^1 using transformations and state the domain and range of f −^1.
- Verify
f −^1 ◦ f
(x) = x for all x in the domain of f and
f ◦ f −^1
(x) = x for all x in the domain of f −^1.
- Graph f and f −^1 on the same set of axes and check the symmetry about the line y = x.
Solution.
- If we identify g(x) = 2x, we see f (x) = g(x − 1) − 3. We pick the points
and (1, 2) on the graph of g along with the horizontal asymptote y = 0 to track through the transformations. By Theorem 1.7 we first add 1 to the x-coordinates of the points on the graph of g (shifting g to the right 1 unit) to get
, (1, 1) and (2, 2). The horizontal asymptote remains y = 0. Next, we subtract 3 from the y-coordinates, shifting the graph down 3 units. We get the points
, (1, −2) and (2, −1) with the horizontal asymptote now at y = −3. Connecting the dots in the order and manner as they were on the graph of g, we get the graph below. We see that the domain of f is the same as g, namely (−∞, ∞), but that the range of f is (− 3 , ∞).
x
y
− 3 − 2 − 1 1 2 3 4 − 3
− 2
− 1
1
2
3
4
5
6
7
y = h(x) = 2x^ −−−−−−−−−−−−→
x
y
− 3 − 2 − 1 1 2 3 4 − 2
− 1
1
2
3
4
5
6
7
y = f (x) = 2x−^1 − 3
6.1 Introduction to Exponential and Logarithmic Functions 427
- The graph of f passes the Horizontal Line Test so f is one-to-one, hence invertible. To find a formula for f −^1 (x), we normally set y = f (x), interchange the x and y, then proceed to solve for y. Doing so in this situation leads us to the equation x = 2y−^1 − 3. We have yet to discuss how to solve this kind of equation, so we will attempt to find the formula for f −^1 from a procedural perspective. If we break f (x) = 2x−^1 − 3 into a series of steps, we find f takes an input x and applies the steps
(a) subtract 1 (b) put as an exponent on 2 (c) subtract 3
Clearly, to undo subtracting 1, we will add 1, and similarly we undo subtracting 3 by adding
- How do we undo the second step? The answer is we use the logarithm. By definition, log 2 (x) undoes exponentiation by 2. Hence, f −^1 should
(a) add 3 (b) take the logarithm base 2 (c) add 1
In symbols, f −^1 (x) = log 2 (x + 3) + 1.
- To graph f −^1 (x) = log 2 (x + 3) + 1 using transformations, we start with j(x) = log 2 (x). We track the points
2 ,^ −^1
, (1, 0) and (2, 1) on the graph of j along with the vertical asymptote x = 0 through the transformations using Theorem 1.7. Since f −^1 (x) = j(x + 3) + 1, we first subtract 3 from each of the x values (including the vertical asymptote) to obtain
(− 2 , 0) and (− 1 , 1) with a vertical asymptote x = −3. Next, we add 1 to the y values on the graph and get
, (− 2 , 1) and (− 1 , 2). If you are experiencing d´ej`a vu, there is a good reason for it but we leave it to the reader to determine the source of this uncanny familiarity. We obtain the graph below. The domain of f −^1 is (− 3 , ∞), which matches the range of f , and the range of f −^1 is (−∞, ∞), which matches the domain of f.
x
y
− 3
− 2
− 1
1
2
3
4
− 3 − 2 − 1 1 2 3 4 5 6 7 8
y = j(x) = log 2 (x) −−−−−−−−−−−−→
x
y
− 3
− 2
− 1
1
2
3
4
− 2 − 1 1 2 3 4 5 6 7 8
y = f −^1 (x) = log 2 (x + 3) + 1
- We now verify that f (x) = 2x−^1 − 3 and f −^1 (x) = log 2 (x + 3) + 1 satisfy the composition requirement for inverses. For all real numbers x,
6.1 Introduction to Exponential and Logarithmic Functions 429
6.1.1 Exercises
In Exercises 1 - 15, use the property: ba^ = c if and only if logb(c) = a from Theorem 6.2 to rewrite the given equation in the other form. That is, rewrite the exponential equations as logarithmic equations and rewrite the logarithmic equations as exponential equations.
2^3 = 8 2. 5−^3 = 1251 3. 4^5 /^2 = 32
3
25
= 52 6. 10−^3 = 0.^001
- e^0 = 1 8. log 5 (25) = 2 9. log 25 (5) = (^12)
- log 3
81
= − 4 11. log (^43)
4
= − 1 12. log(100) = 2
- log(0.1) = − 1 14. ln(e) = 1 (^) 15. ln
√^1 e
In Exercises 16 - 42, evaluate the expression.
- log 3 (27) 17. log 6 (216) 18. log 2 (32)
- log 6
36
- log 8 (4) 21. log 36 (216)
- log 1 5 (625) 23. log 1 6 (216) 24. log 36 (36)
- log
1000000
- log(0.01) 27. ln
e^3
- log 4 (8) 29. log 6 (1) 30. log 13
- log 36
- 7log^7 (3)^ 33. 36log^36 (216)
- log 36
- ln
e^5
- log
- log
- ln
√^1 e
- log 5
3 log^3 (5)
- log
eln(100)
- log 2
3 −^ log^3 (2)
- ln
42 6 log(1)
In Exercises 43 - 57, find the domain of the function.
- f (x) = ln(x^2 + 1) 44. f (x) = log 7 (4x + 8)
- f (x) = ln(4x − 20) 46. f (x) = log
x^2 + 9x + 18
430 Exponential and Logarithmic Functions
- f (x) = log
x + 2 x^2 − 1
- f (x) = log
x^2 + 9x + 18 4 x − 20
- f (x) = ln(7 − x) + ln(x − 4) 50. f (x) = ln(4x − 20) + ln
x^2 + 9x + 18
- f (x) = log
x^2 + x + 1
- f (x) = 4
log 4 (x)
- f (x) = log 9 (|x + 3| − 4) 54. f (x) = ln(
x − 4 − 3)
- f (x) =
3 − log 5 (x) 56.^ f^ (x) =
− 1 − x log 12 (x)
- f (x) = ln(− 2 x^3 − x^2 + 13x − 6)
In Exercises 58 - 63, sketch the graph of y = g(x) by starting with the graph of y = f (x) and using transformations. Track at least three points of your choice and the horizontal asymptote through the transformations. State the domain and range of g.
- f (x) = 2x, g(x) = 2x^ − (^1) 59. f (x) =
3
)x , g(x) =
3
)x− 1
- f (x) = 3x, g(x) = 3−x^ + 2 61. f (x) = 10x, g(x) = 10 x+1 2 − 20
- f (x) = ex, g(x) = 8 − e−x^ 63. f (x) = ex, g(x) = 10e−^0.^1 x
In Exercises 64 - 69, sketch the graph of y = g(x) by starting with the graph of y = f (x) and using transformations. Track at least three points of your choice and the vertical asymptote through the transformations. State the domain and range of g.
- f (x) = log 2 (x), g(x) = log 2 (x + 1) 65. f (x) = log 1 3 (x), g(x) = log 1 3 (x) + 1
- f (x) = log 3 (x), g(x) = − log 3 (x − 2) 67. f (x) = log(x), g(x) = 2 log(x + 20) − 1
- f (x) = ln(x), g(x) = − ln(8 − x) 69. f (x) = ln(x), g(x) = −10 ln
( (^) x 10
- Verify that each function in Exercises 64 - 69 is the inverse of the corresponding function in Exercises 58 - 63. (Match up #58 and #64, and so on.)
In Exercises 71 - 74, find the inverse of the function from the ‘procedural perspective’ discussed in Example 6.1.5 and graph the function and its inverse on the same set of axes.
- f (x) = 3x+2^ − 4 72. f (x) = log 4 (x − 1)
- f (x) = − 2 −x^ + 1 74. f (x) = 5 log(x) − 2
432 Exponential and Logarithmic Functions
(b) Damage to your hearing can start with short term exposure to sound levels around 115 decibels. What intensity I is needed to produce this level? (c) Compute L(1). How does this compare with the threshold of pain which is around 140 decibels?
- The pH of a solution is a measure of its acidity or alkalinity. Specifically, pH = − log[H+] where [H+] is the hydrogen ion concentration in moles per liter. A solution with a pH less than 7 is an acid, one with a pH greater than 7 is a base (alkaline) and a pH of 7 is regarded as neutral.
(a) The hydrogen ion concentration of pure water is [H+] = 10−^7. Find its pH. (b) Find the pH of a solution with [H+] = 6. 3 × 10 −^13. (c) The pH of gastric acid (the acid in your stomach) is about 0.7. What is the corresponding hydrogen ion concentration?
- Show that logb 1 = 0 and logb b = 1 for every b > 0 , b 6 = 1.
- (Crazy bonus question) Without using your calculator, determine which is larger: eπ^ or πe.
6.1 Introduction to Exponential and Logarithmic Functions 433
6.1.2 Answers
- log 2 (8) = 3 2. log 5
125
= − 3 3. log 4 (32) = (^52)
- log 1 3 (9) = − 2 5. log 4 25
2
= − 12 6. log(0.001) = − 3
- ln(1) = 0 8. 5^2 = 25 9. (25)
1 (^2) = 5
10. 3−^4 = 811 11.
3
= 34 12. 10^2 = 100
- 10−^1 = 0. 1 14. e^1 = e (^) 15. e−^ (^12) = √^1 e
- log 3 (27) = 3 17. log 6 (216) = 3 18. log 2 (32) = 5
- log 6
36
= − 2 20. log 8 (4) = 23 21. log 36 (216) = (^32)
- log 15 (625) = − 4 23. log 16 (216) = − 3 24. log 36 (36) = 1
- log 10000001 = − 6 26. log(0.01) = − 2 27. ln
e^3
- log 4 (8) = 32 29. log 6 (1) = 0 30. log 13
- log 36
= 14 32. 7log^7 (3)^ = 3 33. 36log^36 (216)^ = 216
- log 36
= 216 35. ln(e^5 ) = 5 (^) 36. log
- log
= 53 38. ln
√^1 e
= − 12 39. log 5
3 log^3
- log
eln(100)
= 2 41. log 2
3 −^ log^3 (2)
= − 1 42. ln
42 6 log(1)
52. [1, ∞) 53. (−∞, −7) ∪ (1, ∞) 54. (13, ∞)
- (0, 125) ∪ (125, ∞) 56. No domain 57. (−∞, −3) ∪
2 ,^2
6.1 Introduction to Exponential and Logarithmic Functions 435
- Domain of g: (− 1 , ∞) Range of g: (−∞, ∞)
y
x
− 3
− 2
− 1
V.A. x = − 1
1
2
3
1 2 3 4 5 6 7 8
y = g(x) = log 2 (x + 1)
- Domain of g: (0, ∞) Range of g: (−∞, ∞)
y
x
− 3
− 2
− 1
1
2
3
1 2 3 4 5 6 7 8 9
y = g(x) = log 13 (x) + 1
- Domain of g: (2, ∞) Range of g: (−∞, ∞)
y
x
V.A. x = 2
− 3
− 2
− 1
1
2
3
1 2 3 4 5 6 7 8 9 10 11
y = g(x) = − log 3 (x − 2)
- Domain of g: (− 20 , ∞) Range of g: (−∞, ∞)
y
x
− 3 V.A. x = − 20
− 2
1
2
3
− 10 10 20 30 40 50 60 70 80 90 100
y = g(x) = 2 log(x + 20) − 1
- Domain of g: (−∞, 8) Range of g:(−∞, ∞) y
x
− 3 V.A. x = 8
− 2
− 1
1
2
3
1 2 3 4 5 6 7 8
y = g(x) = − ln(8 − x)
- Domain of g: (0, ∞) Range of g: (−∞, ∞)
y
− 10 x
10
20
30
10 20 30 40 50 60 70 80
y = g(x) = −10 ln (^10 x^ )
436 Exponential and Logarithmic Functions
- f (x) = 3x+2^ − 4 f −^1 (x) = log 3 (x + 4) − 2
x
y
y = f (x) = 3x+2^ − 4 y = f −^1 (x) = log 3 (x + 4) − 2
− 4 − 3 − 2 − 1 1 2 3 4 5 6
− 4
− 3
− 2
− 1
1
2
3
4
5
6
- f (x) = log 4 (x − 1) f −^1 (x) = 4x^ + 1
x
y
y = f (x) = log 4 (x − 1) y = f −^1 (x) = 4x^ + 1
− 2 − 1 1 2 3 4 5 6 − 2
− 1
1
2
3
4
5
6
- f (x) = − 2 −x^ + 1 f −^1 (x) = − log 2 (1 − x)
x
y
y = f (x) = − 2 −x^ + 1 y = f −^1 (x) = − log 2 (1 − x)
− 2 − 1 1 2
− 2
− 1
1
2
- f (x) = 5 log(x) − 2 f −^1 (x) = 10
x+ 5
x
y
y = f (x) = 5 log(x) − 2 y = f −^1 (x) = 10 x+2^5
− 4 − 3 − 2 − 1 1 2 3 4 5
− 4
− 3
− 2
− 1
1
2
3
4
5
- (a) M (0.001) = log
- 001
= log(1) = 0. (b) M (80, 000) = log
80 , 000
- 001
= log(80, 000 , 000) ≈ 7 .9.
- (a) L(10−^6 ) = 60 decibels. (b) I = 10−.^5 ≈ 0 .316 watts per square meter. (c) Since L(1) = 120 decibels and L(100) = 140 decibels, a sound with intensity level 140 decibels has an intensity 100 times greater than a sound with intensity level 120 decibels.
- (a) The pH of pure water is 7. (b) If [H+] = 6. 3 × 10 −^13 then the solution has a pH of 12.2. (c) [H+] = 10−^0.^7 ≈ .1995 moles per liter.
