In this section we examine exponential and logarithmic functions. We use the properties of these functions to solve equations involving exponential or logarithmic terms, and we study the meaning and importance of the number e . e . We also define hyperbolic and inverse hyperbolic functions, which involve combinations of exponential and logarithmic functions. (Note that we present alternative definitions of exponential and logarithmic functions in the chapter Applications of Integrations, and prove that the functions have the same properties with either definition.)
Exponential functions arise in many applications. One common example is population growth .
For example, if a population starts with P 0 P 0 individuals and then grows at an annual rate of 2 % , 2 % , its population after 1 year is
P ( 1 ) = P 0 + 0.02 P 0 = P 0 ( 1 + 0.02 ) = P 0 ( 1.02 ) . P ( 1 ) = P 0 + 0.02 P 0 = P 0 ( 1 + 0.02 ) = P 0 ( 1.02 ) .
Its population after 2 years is
P ( 2 ) = P ( 1 ) + 0.02 P ( 1 ) = P ( 1 ) ( 1.02 ) = P 0 ( 1.02 ) 2 . P ( 2 ) = P ( 1 ) + 0.02 P ( 1 ) = P ( 1 ) ( 1.02 ) = P 0 ( 1.02 ) 2 .
In general, its population after t t years is
P ( t ) = P 0 ( 1.02 ) t , P ( t ) = P 0 ( 1.02 ) t ,which is an exponential function. More generally, any function of the form f ( x ) = b x , f ( x ) = b x , where b > 0 , b ≠ 1 , b > 0 , b ≠ 1 , is an exponential function with base b b and exponent x. Exponential functions have constant bases and variable exponents. Note that a function of the form f ( x ) = x b f ( x ) = x b for some constant b b is not an exponential function but a power function.
To see the difference between an exponential function and a power function, we compare the functions y = x 2 y = x 2 and y = 2 x . y = 2 x . In Table 1.10, we see that both 2 x 2 x and x 2 x 2 approach infinity as x → ∞ . x → ∞ . Eventually, however, 2 x 2 x becomes larger than x 2 x 2 and grows more rapidly as x → ∞ . x → ∞ . In the opposite direction, as x → − ∞ , x 2 → ∞ , x → − ∞ , x 2 → ∞ , whereas 2 x → 0 . 2 x → 0 . The line y = 0 y = 0 is a horizontal asymptote for y = 2 x . y = 2 x .
| x x | −3 −3 | −2 −2 | −1 −1 | 0 0 | 1 1 | 2 2 | 3 3 | 4 4 | 5 5 | 6 6 |
| x 2 x 2 | 9 9 | 4 4 | 1 1 | 0 0 | 1 1 | 4 4 | 9 9 | 16 16 | 25 25 | 36 36 |
| 2 x 2 x | 1 / 8 1 / 8 | 1 / 4 1 / 4 | 1 / 2 1 / 2 | 1 1 | 2 2 | 4 4 | 8 8 | 16 16 | 32 32 | 64 64 |
In Figure 1.43, we graph both y = x 2 y = x 2 and y = 2 x y = 2 x to show how the graphs differ.
Figure 1.43 Both 2 x 2 x and x 2 x 2 approach infinity as x → ∞ , x → ∞ , but 2 x 2 x grows more rapidly than x 2 . x 2 . As x → − ∞ , x 2 → ∞ , x → − ∞ , x 2 → ∞ , whereas 2 x → 0 . 2 x → 0 .
Recall the properties of exponents: If x x is a positive integer, then we define b x = b · b ⋯ b b x = b · b ⋯ b (with x x factors of b ) . b ) . If x x is a negative integer, then x = − y x = − y for some positive integer y , y , and we define b x = b − y = 1 / b y . b x = b − y = 1 / b y . Also, b 0 b 0 is defined to be 1 . 1 . If x x is a rational number, then x = p / q , x = p / q , where p p and q q are integers and b x = b p / q = b p q . b x = b p / q = b p q . For example, 9 3 / 2 = 9 3 = 27 . 9 3 / 2 = 9 3 = 27 . However, how is b x b x defined if x x is an irrational number? For example, what do we mean by 2 2 ? 2 2 ? This is too complex a question for us to answer fully right now; however, we can make an approximation. In Table 1.11, we list some rational numbers approaching 2 , 2 , and the values of 2 x 2 x for each rational number x x are presented as well. We claim that if we choose rational numbers x x getting closer and closer to 2 , 2 , the values of 2 x 2 x get closer and closer to some number L . L . We define that number L L to be 2 2 . 2 2 .
| x x | 1.4 1.4 | 1.41 1.41 | 1.414 1.414 | 1.4142 1.4142 | 1.41421 1.41421 | 1.414213 1.414213 |
| 2 x 2 x | 2.639 2.639 | 2.65737 2.65737 | 2.66475 2.66475 | 2.665119 2.665119 | 2.665138 2.665138 | 2.665143 2.665143 |
Suppose a particular population of bacteria is known to double in size every 4 4 hours. If a culture starts with 1000 1000 bacteria, the number of bacteria after 4 4 hours is n ( 4 ) = 1000 · 2 . n ( 4 ) = 1000 · 2 . The number of bacteria after 8 8 hours is n ( 8 ) = n ( 4 ) · 2 = 1000 · 2 2 . n ( 8 ) = n ( 4 ) · 2 = 1000 · 2 2 . In general, the number of bacteria after 4 m 4 m hours is n ( 4 m ) = 1000 · 2 m . n ( 4 m ) = 1000 · 2 m . Letting t = 4 m , t = 4 m , we see that the number of bacteria after t t hours is n ( t ) = 1000 · 2 t / 4 . n ( t ) = 1000 · 2 t / 4 . Find the number of bacteria after 6 6 hours, 10 10 hours, and 24 24 hours.
The number of bacteria after 6 hours is given by n ( 6 ) = 1000 · 2 6 / 4 ≈ 2828 n ( 6 ) = 1000 · 2 6 / 4 ≈ 2828 bacteria. The number of bacteria after 10 10 hours is given by n ( 10 ) = 1000 · 2 10 / 4 ≈ 5657 n ( 10 ) = 1000 · 2 10 / 4 ≈ 5657 bacteria. The number of bacteria after 24 24 hours is given by n ( 24 ) = 1000 · 2 6 = 64,000 n ( 24 ) = 1000 · 2 6 = 64,000 bacteria.
Given the exponential function f ( x ) = 100 · 3 x / 2 , f ( x ) = 100 · 3 x / 2 , evaluate f ( 4 ) f ( 4 ) and f ( 10 ) . f ( 10 ) .
Go to Population Balance for another example of exponential population growth.
For any base b > 0 , b ≠ 1 , b > 0 , b ≠ 1 , the exponential function f ( x ) = b x f ( x ) = b x is defined for all real numbers x x and b x > 0 . b x > 0 . Therefore, the domain of f ( x ) = b x f ( x ) = b x is ( − ∞ , ∞ ) ( − ∞ , ∞ ) and the range is ( 0 , ∞ ) . ( 0 , ∞ ) . To graph b x , b x , we note that for b > 1 , b x b > 1 , b x is increasing on ( − ∞ , ∞ ) ( − ∞ , ∞ ) and b x → ∞ b x → ∞ as x → ∞ , x → ∞ , whereas b x → 0 b x → 0 as x → − ∞ . x → − ∞ . On the other hand, if 0 < b < 1 , f ( x ) = b x 0 < b < 1 , f ( x ) = b x is decreasing on ( − ∞ , ∞ ) ( − ∞ , ∞ ) and b x → 0 b x → 0 as x → ∞ x → ∞ whereas b x → ∞ b x → ∞ as x → − ∞ x → − ∞ (Figure 1.44).
Figure 1.44 If b > 1 , b > 1 , then b x b x is increasing on ( − ∞ , ∞ ) . ( − ∞ , ∞ ) . If 0 < b < 1 , 0 < b < 1 , then b x b x is decreasing on ( − ∞ , ∞ ) . ( − ∞ , ∞ ) .
Visit this site for more exploration of the graphs of exponential functions.
Note that exponential functions satisfy the general laws of exponents. To remind you of these laws, we state them as rules.
For any constants a > 0 , b > 0 , a > 0 , b > 0 , and for all x and y,
Use the laws of exponents to simplify each of the following expressions.
( 2 x 2 / 3 ) 3 ( 4 x −1 / 3 ) 2 = 2 3 ( x 2 / 3 ) 3 4 2 ( x −1 / 3 ) 2 = 8 x 2 16 x −2 / 3 = x 2 x 2 / 3 2 = x 8 / 3 2 . ( 2 x 2 / 3 ) 3 ( 4 x −1 / 3 ) 2 = 2 3 ( x 2 / 3 ) 3 4 2 ( x −1 / 3 ) 2 = 8 x 2 16 x −2 / 3 = x 2 x 2 / 3 2 = x 8 / 3 2 .
( x 3 y −1 ) 2 ( x y 2 ) −2 = ( x 3 ) 2 ( y −1 ) 2 x −2 ( y 2 ) −2 = x 6 y −2 x −2 y −4 = x 6 x 2 y −2 y 4 = x 8 y 2 . ( x 3 y −1 ) 2 ( x y 2 ) −2 = ( x 3 ) 2 ( y −1 ) 2 x −2 ( y 2 ) −2 = x 6 y −2 x −2 y −4 = x 6 x 2 y −2 y 4 = x 8 y 2 .
Use the laws of exponents to simplify ( 6 x −3 y 2 ) / ( 12 x −4 y 5 ) . ( 6 x −3 y 2 ) / ( 12 x −4 y 5 ) .
A special type of exponential function appears frequently in real-world applications. To describe it, consider the following example of exponential growth, which arises from compounding interest in a savings account. Suppose a person invests P P dollars in a savings account with an annual interest rate r , r , compounded annually. The amount of money after 1 year is
A ( 1 ) = P + r P = P ( 1 + r ) . A ( 1 ) = P + r P = P ( 1 + r ) .The amount of money after 2 2 years is
A ( 2 ) = A ( 1 ) + r A ( 1 ) = P ( 1 + r ) + r P ( 1 + r ) = P ( 1 + r ) 2 . A ( 2 ) = A ( 1 ) + r A ( 1 ) = P ( 1 + r ) + r P ( 1 + r ) = P ( 1 + r ) 2 .
More generally, the amount after t t years is
A ( t ) = P ( 1 + r ) t . A ( t ) = P ( 1 + r ) t .If the money is compounded 2 times per year, the amount of money after half a year is
A ( 1 2 ) = P + ( r 2 ) P = P ( 1 + ( r 2 ) ) . A ( 1 2 ) = P + ( r 2 ) P = P ( 1 + ( r 2 ) ) .The amount of money after 1 1 year is
A ( 1 ) = A ( 1 2 ) + ( r 2 ) A ( 1 2 ) = P ( 1 + r 2 ) + r 2 ( P ( 1 + r 2 ) ) = P ( 1 + r 2 ) 2 . A ( 1 ) = A ( 1 2 ) + ( r 2 ) A ( 1 2 ) = P ( 1 + r 2 ) + r 2 ( P ( 1 + r 2 ) ) = P ( 1 + r 2 ) 2 .
After t t years, the amount of money in the account is
A ( t ) = P ( 1 + r 2 ) 2 t . A ( t ) = P ( 1 + r 2 ) 2 t .More generally, if the money is compounded n n times per year, the amount of money in the account after t t years is given by the function
A ( t ) = P ( 1 + r n ) n t . A ( t ) = P ( 1 + r n ) n t .What happens as n → ∞ ? n → ∞ ? To answer this question, we let m = n / r m = n / r and write
( 1 + r n ) n t = ( 1 + 1 m ) m r t , ( 1 + r n ) n t = ( 1 + 1 m ) m r t ,and examine the behavior of ( 1 + 1 / m ) m ( 1 + 1 / m ) m as m → ∞ , m → ∞ , using a table of values (Table 1.12).
| m m | 10 10 | 100 100 | 1000 1000 | 10,000 10,000 | 100,000 100,000 | 1,000,000 1,000,000 |
| ( 1 + 1 m ) m ( 1 + 1 m ) m | 2.5937 2.5937 | 2.7048 2.7048 | 2.71692 2.71692 | 2.71815 2.71815 | 2.718268 2.718268 | 2.718280 2.718280 |
Looking at this table, it appears that ( 1 + 1 / m ) m ( 1 + 1 / m ) m is approaching a number between 2.7 2.7 and 2.8 2.8 as m → ∞ . m → ∞ . In fact, ( 1 + 1 / m ) m ( 1 + 1 / m ) m does approach some number as m → ∞ . m → ∞ . We call this number e e . To six decimal places of accuracy,
e ≈ 2.718282 . e ≈ 2.718282 .The letter e e was first used to represent this number by the Swiss mathematician Leonhard Euler during the 1720s. Although Euler did not discover the number, he showed many important connections between e e and logarithmic functions. We still use the notation e e today to honor Euler’s work because it appears in many areas of mathematics and because we can use it in many practical applications.
Returning to our savings account example, we can conclude that if a person puts P P dollars in an account at an annual interest rate r , r , compounded continuously, then A ( t ) = P e r t . A ( t ) = P e r t . This function may be familiar. Since functions involving base e e arise often in applications, we call the function f ( x ) = e x f ( x ) = e x the natural exponential function . Not only is this function interesting because of the definition of the number e , e , but also, as discussed next, its graph has an important property.
Since e > 1 , e > 1 , we know e x e x is increasing on ( − ∞ , ∞ ) . ( − ∞ , ∞ ) . In Figure 1.45, we show a graph of f ( x ) = e x f ( x ) = e x along with a tangent line to the graph of at x = 0 . x = 0 . We give a precise definition of tangent line in the next chapter; but, informally, we say a tangent line to a graph of f f at x = a x = a is a line that passes through the point ( a , f ( a ) ) ( a , f ( a ) ) and has the same “slope” as f f at that point . . The function f ( x ) = e x f ( x ) = e x is the only exponential function b x b x with tangent line at x = 0 x = 0 that has a slope of 1. As we see later in the text, having this property makes the natural exponential function the most simple exponential function to use in many instances.
Figure 1.45 The graph of f ( x ) = e x f ( x ) = e x has a tangent line with slope 1 1 at x = 0 . x = 0 .
Suppose $ 500 $ 500 is invested in an account at an annual interest rate of r = 5.5 % , r = 5.5 % , compounded continuously.
A ( 10 ) = 500 e 0.055 · 10 = 500 e 0.55 ≈ $ 866.63 . A ( 10 ) = 500 e 0.055 · 10 = 500 e 0.55 ≈ $ 866.63 .
A ( 20 ) = 500 e 0.055 · 20 = 500 e 1.1 ≈ $ 1 , 502.08 . A ( 20 ) = 500 e 0.055 · 20 = 500 e 1.1 ≈ $ 1 , 502.08 .
If $ 750 $ 750 is invested in an account at an annual interest rate of 4 % , 4 % , compounded continuously, find a formula for the amount of money in the account after t t years. Find the amount of money after 30 30 years.
Using our understanding of exponential functions, we can discuss their inverses, which are the logarithmic functions. These come in handy when we need to consider any phenomenon that varies over a wide range of values, such as pH in chemistry or decibels in sound levels.
The exponential function f ( x ) = b x f ( x ) = b x is one-to-one, with domain ( − ∞ , ∞ ) ( − ∞ , ∞ ) and range ( 0 , ∞ ) . ( 0 , ∞ ) . Therefore, it has an inverse function, called the logarithmic function with base b . b . For any b > 0 , b ≠ 1 , b > 0 , b ≠ 1 , the logarithmic function with base b, denoted log b , log b , has domain ( 0 , ∞ ) ( 0 , ∞ ) and range ( − ∞ , ∞ ) , ( − ∞ , ∞ ) , and satisfies
log b ( x ) = y if and only if b y = x . log b ( x ) = y if and only if b y = x .log 2 ( 8 ) = 3 since 2 3 = 8 , log 10 ( 1 100 ) = −2 since 10 −2 = 1 10 2 = 1 100 , log b ( 1 ) = 0 since b 0 = 1 for any base b > 0 . log 2 ( 8 ) = 3 since 2 3 = 8 , log 10 ( 1 100 ) = −2 since 10 −2 = 1 10 2 = 1 100 , log b ( 1 ) = 0 since b 0 = 1 for any base b > 0 .
Furthermore, since y = log b ( x ) y = log b ( x ) and y = b x y = b x are inverse functions,
log b ( b x ) = x and b log b ( x ) = x . log b ( b x ) = x and b log b ( x ) = x .The most commonly used logarithmic function is the function log e . log e . Since this function uses natural e e as its base, it is called the natural logarithm . Here we use the notation ln ( x ) ln ( x ) or ln x ln x to mean log e ( x ) . log e ( x ) . For example,
ln ( e ) = log e ( e ) = 1 , ln ( e 3 ) = log e ( e 3 ) = 3 , ln ( 1 ) = log e ( 1 ) = 0 . ln ( e ) = log e ( e ) = 1 , ln ( e 3 ) = log e ( e 3 ) = 3 , ln ( 1 ) = log e ( 1 ) = 0 .
Since the functions f ( x ) = e x f ( x ) = e x and g ( x ) = ln ( x ) g ( x ) = ln ( x ) are inverses of each other,
ln ( e x ) = x and e ln x = x , ln ( e x ) = x and e ln x = x ,and their graphs are symmetric about the line y = x y = x (Figure 1.46).
Figure 1.46 The functions y = e x y = e x and y = ln ( x ) y = ln ( x ) are inverses of each other, so their graphs are symmetric about the line y = x . y = x .
At this site you can see an example of a base-10 logarithmic scale.
In general, for any base b > 0 , b ≠ 1 , b > 0 , b ≠ 1 , the function g ( x ) = log b ( x ) g ( x ) = log b ( x ) is symmetric about the line y = x y = x with the function f ( x ) = b x . f ( x ) = b x . Using this fact and the graphs of the exponential functions, we graph functions log b log b for several values of b > 1 b > 1 (Figure 1.47).
Figure 1.47 Graphs of y = log b ( x ) y = log b ( x ) are depicted for b = 2 , e , 10 . b = 2 , e , 10 .
Before solving some equations involving exponential and logarithmic functions, let’s review the basic properties of logarithms.
If a , b , c > 0 , b ≠ 1 , a , b , c > 0 , b ≠ 1 , and r r is any real number, then
1. log b ( a c ) = log b ( a ) + log b ( c ) (Product property) 2. log b ( a c ) = log b ( a ) − log b ( c ) (Quotient property) 3. log b ( a r ) = r log b ( a ) (Power property) 1. log b ( a c ) = log b ( a ) + log b ( c ) (Product property) 2. log b ( a c ) = log b ( a ) − log b ( c ) (Quotient property) 3. log b ( a r ) = r log b ( a ) (Power property)
Solve each of the following equations for x . x .
Solve e 2 x / ( 3 + e 2 x ) = 1 / 2 . e 2 x / ( 3 + e 2 x ) = 1 / 2 .
Solve each of the following equations for x . x .
log 10 x + log 10 x = log 10 x x = log 10 x 3 / 2 = 3 2 log 10 x . log 10 x + log 10 x = log 10 x x = log 10 x 3 / 2 = 3 2 log 10 x .
Solve ln ( x 3 ) − 4 ln ( x ) = 1 . ln ( x 3 ) − 4 ln ( x ) = 1 .
When evaluating a logarithmic function with a calculator, you may have noticed that the only options are log 10 log 10 or log, called the common logarithm , or ln, which is the natural logarithm. However, exponential functions and logarithm functions can be expressed in terms of any desired base b . b . If you need to use a calculator to evaluate an expression with a different base, you can apply the change-of-base formulas first. Using this change of base, we typically write a given exponential or logarithmic function in terms of the natural exponential and natural logarithmic functions.
Let a > 0 , b > 0 , a > 0 , b > 0 , and a ≠ 1 , b ≠ 1 . a ≠ 1 , b ≠ 1 .
For the first change-of-base formula, we begin by making use of the power property of logarithmic functions. We know that for any base b > 0 , b ≠ 1 , log b ( a x ) = x log b a . b > 0 , b ≠ 1 , log b ( a x ) = x log b a . Therefore,
b log b ( a x ) = b x log b a . b log b ( a x ) = b x log b a .In addition, we know that b x b x and log b ( x ) log b ( x ) are inverse functions. Therefore,
b log b ( a x ) = a x . b log b ( a x ) = a x .Combining these last two equalities, we conclude that a x = b x log b a . a x = b x log b a .
To prove the second property, we show that
( log b a ) · ( log a x ) = log b x . ( log b a ) · ( log a x ) = log b x .Let u = log b a , v = log a x , u = log b a , v = log a x , and w = log b x . w = log b x . We will show that u · v = w . u · v = w . By the definition of logarithmic functions, we know that b u = a , a v = x , b u = a , a v = x , and b w = x . b w = x . From the previous equations, we see that
b u v = ( b u ) v = a v = x = b w . b u v = ( b u ) v = a v = x = b w .Therefore, b u v = b w . b u v = b w . Since exponential functions are one-to-one, we can conclude that u · v = w . u · v = w .
Use a calculating utility to evaluate log 3 7 log 3 7 with the change-of-base formula presented earlier.
Use the second equation with a = 3 a = 3 and e = 3 : e = 3 :
log 3 7 = ln 7 ln 3 ≈ 1.77124 . log 3 7 = ln 7 ln 3 ≈ 1.77124 .
Use the change-of-base formula and a calculating utility to evaluate log 4 6 . log 4 6 .
In 1935, Charles Richter developed a scale (now known as the Richter scale ) to measure the magnitude of an earthquake . The scale is a base-10 logarithmic scale, and it can be described as follows: Consider one earthquake with magnitude R 1 R 1 on the Richter scale and a second earthquake with magnitude R 2 R 2 on the Richter scale. Suppose R 1 > R 2 , R 1 > R 2 , which means the earthquake of magnitude R 1 R 1 is stronger, but how much stronger is it than the other earthquake? A way of measuring the intensity of an earthquake is by using a seismograph to measure the amplitude of the earthquake waves. If A 1 A 1 is the amplitude measured for the first earthquake and A 2 A 2 is the amplitude measured for the second earthquake, then the amplitudes and magnitudes of the two earthquakes satisfy the following equation:
R 1 − R 2 = log 10 ( A 1 A 2 ) . R 1 − R 2 = log 10 ( A 1 A 2 ) .Consider an earthquake that measures 8 on the Richter scale and an earthquake that measures 7 on the Richter scale. Then,
8 − 7 = log 10 ( A 1 A 2 ) . 8 − 7 = log 10 ( A 1 A 2 ) . log 10 ( A 1 A 2 ) = 1 , log 10 ( A 1 A 2 ) = 1 ,which implies A 1 / A 2 = 10 A 1 / A 2 = 10 or A 1 = 10 A 2 . A 1 = 10 A 2 . Since A 1 A 1 is 10 times the size of A 2 , A 2 , we say that the first earthquake is 10 times as intense as the second earthquake. On the other hand, if one earthquake measures 8 on the Richter scale and another measures 6, then the relative intensity of the two earthquakes satisfies the equation
log 10 ( A 1 A 2 ) = 8 − 6 = 2 . log 10 ( A 1 A 2 ) = 8 − 6 = 2 .Therefore, A 1 = 100 A 2 . A 1 = 100 A 2 . That is, the first earthquake is 100 times more intense than the second earthquake.
How can we use logarithmic functions to compare the relative severity of the magnitude 9 earthquake in Japan in 2011 with the magnitude 7.3 earthquake in Haiti in 2010?
To compare the Japan and Haiti earthquakes, we can use an equation presented earlier:
9 − 7.3 = log 10 ( A 1 A 2 ) . 9 − 7.3 = log 10 ( A 1 A 2 ) .
Therefore, A 1 / A 2 = 10 1.7 , A 1 / A 2 = 10 1.7 , and we conclude that the earthquake in Japan was approximately 50 50 times more intense than the earthquake in Haiti.
