1.5: Exponential and Logarithmic Functions
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- Identify the form of an exponential function.
- Explain the difference between the graphs of xb and bx.
- Recognize the significance of the number e.
- Identify the form of a logarithmic function.
- Explain the relationship between exponential and logarithmic functions.
- Describe how to calculate a logarithm to a different base.
- Identify the hyperbolic functions, their graphs, and basic identities.
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. 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
Exponential functions arise in many applications. One common example is population growth. For example, if a population starts with P0 individuals and then grows at an annual rate of 2%, its population after 1 year is
P(1)=P0+0.02P0=P0(1+0.02)=P0(1.02).
Its population after 2 years is
P(2)=P(1)+0.02P(1)=P(1)(1.02)=P0(1.02)2.
In general, its population after t years is
P(t)=P0(1.02)t,
which is an exponential function. More generally, any function of the form f(x)=bx, where b>0, b≠1, is an exponential function with base b and exponent x. Exponential functions have constant bases and variable exponents. Note that a function of the form f(x)=xb for some constant 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=x2 and y=2x. In Table 1.5.1, we see that both 2x and x2 approach infinity as x→∞. Eventually, however, 2x becomes larger than x2 and grows more rapidly as x→∞. In the opposite direction, as x→−∞, x2→∞, whereas 2x→0. The line y=0 is a horizontal asymptote for y=2x.
x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
---|---|---|---|---|---|---|---|---|---|---|
x2 | 9 | 4 | 1 | 0 | 1 | 4 | 9 | 16 | 25 | 36 |
2x | 1/8 | 1/4 | 1/2 | 1 | 2 | 4 | 8 | 16 | 32 | 64 |
In Figure 1.5.1, we graph both y=x2 and y=2x to show how the graphs differ.

Evaluating Exponential Functions
Recall the properties of exponents: If x is a positive integer, then we define bx=b⋅b⋯b (with x factors of b). If x is a negative integer, then x=−y for some positive integer y, and we define bx=b−y=1/by. Also, b0 is defined to be 1. If x is a rational number, then x=p/q, where p and q are integers and bx=bp/q=q√bp. For example, 93/2=√93=(√9)3=27. However, how is bx defined if x is an irrational number? For example, what do we mean by 2√2? This is too complex a question for us to answer fully right now; however, we can make an approximation.
x | 1.4 | 1.41 | 1.414 | 1.4142 | 1.41421 | 1.414213 |
---|---|---|---|---|---|---|
2x | 2.639 | 2.65737 | 2.66475 | 2.665119 | 2.665138 | 2.665143 |
In Table 1.5.2, we list some rational numbers approaching √2, and the values of 2x for each rational number x are presented as well. We claim that if we choose rational numbers x getting closer and closer to √2, the values of 2x get closer and closer to some number L. We define that number L to be 2√2.
Suppose a particular population of bacteria is known to double in size every 4 hours. If a culture starts with 1000 bacteria, the number of bacteria after 4 hours is n(4)=1000⋅2. The number of bacteria after 8 hours is n(8)=n(4)⋅2=1000⋅22. In general, the number of bacteria after 4m hours is n(4m)=1000⋅2m. Letting t=4m, we see that the number of bacteria after t hours is n(t)=1000⋅2t/4. Find the number of bacteria after 6 hours, 10 hours, and 24 hours.
Solution
The number of bacteria after 6 hours is given by
n(6)=1000⋅26/4≈2828bacteria.
The number of bacteria after 10 hours is given by
n(10)=1000⋅210/4≈5657bacteria.
The number of bacteria after 24 hours is given by n(24)=1000⋅26=64,000 bacteria.
Given the exponential function f(x)=100⋅3x/2, evaluate f(4) and f(10).
- Answer
-
f(4)=900
f(10)=24,300.
Graphing Exponential Functions
For any base b>0, b≠1, the exponential function f(x)=bx is defined for all real numbers x and bx>0. Therefore, the domain of f(x)=bx is (−∞,∞) and the range is (0,∞). To graph bx, we note that for b>1, bx is increasing on (−∞,∞) and bx→∞ as x→∞, whereas bx→0 as x→−∞. On the other hand, if 0<b<1, f(x)=bx is decreasing on (−∞,∞) and bx→0 as x→∞ whereas bx→∞ as x→−∞ (Figure 1.5.2).

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, and for all x and y,
- bx⋅by=bx+y
- bxby=bx−y
- (bx)y=bxy
- (ab)x=axbx
- axbx=(ab)x
Use the laws of exponents to simplify each of the following expressions.
- (2x2/3)3(4x−1/3)2
- (x3y−1)2(xy2)−2
Soution
a. We can simplify as follows:
(2x2/3)3(4x−1/3)2=23(x2/3)342(x−1/3)2=8x216x−2/3=x2x2/32=x8/32.
b. We can simplify as follows:
(x3y−1)2(xy2)−2=(x3)2(y−1)2x−2(y2)−2=x6y−2x−2y−4=x6x2y−2y4=x8y2.
Use the laws of exponents to simplify 6x−3y212x−4y5.
- Hint
-
xa/xb=xa−b
- Answer
-
x/(2y3)
The Number e
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 dollars in a savings account with an annual interest rate r, compounded annually. The amount of money after 1 year is
A(1)=P+rP=P(1+r).
The amount of money after 2 years is
A(2)=A(1)+rA(1)=P(1+r)+rP(1+r)=P(1+r)2.
More generally, the amount after t years is
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(12)=P+(r2)P=P(1+(r2)).
The amount of money after 1 year is
A(1)=A(12)+(r2)A(12)=P(1+r2)+r2((P(1+r2))=P(1+r2)2.
After t years, the amount of money in the account is
A(t)=P(1+r2)2t.
More generally, if the money is compounded n times per year, the amount of money in the account after t years is given by the function
A(t)=P(1+rn)nt.
What happens as n→∞? To answer this question, we let m=n/r and write
(1+rn)nt=(1+1m)mrt,
and examine the behavior of (1+1/m)m as m→∞, using a table of values (Table 1.5.3).
m | 10 | 100 | 1000 | 10,000 | 100,000 | 1,000,000 |
---|---|---|---|---|---|---|
(1+1m)m | 2.5937 | 2.7048 | 2.71692 | 2.71815 | 2.718268 | 2.718280 |
Looking at this table, it appears that (1+1/m)m is approaching a number between 2.7 and 2.8 as m→∞. In fact, (1+1/m)m does approach some number as m→∞. We call this number e. To six decimal places of accuracy,
e≈2.718282.
The letter 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 and logarithmic functions. We still use the notation 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 dollars in an account at an annual interest rate r, compounded continuously, then A(t)=Pert. This function may be familiar. Since functions involving base e arise often in applications, we call the function f(x)=ex the natural exponential function. Not only is this function interesting because of the definition of the number e, but also, as discussed next, its graph has an important property.
Since e>1, we know f(x)=ex is increasing on (−∞,∞). In Figure 1.5.3, we show a graph of f(x)=ex along with a tangent line to the graph of f at 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 at x=a is a line that passes through the point (a,f(a)) and has the same “slope” as f at that point . The function f(x)=ex is the only exponential function bx with tangent line at 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.

Suppose $500 is invested in an account at an annual interest rate of r=5.5%, compounded continuously.
- Let t denote the number of years after the initial investment and A(t) denote the amount of money in the account at time t. Find a formula for A(t).
- Find the amount of money in the account after 10 years and after 20 years.
Solution
a. If P dollars are invested in an account at an annual interest rate r, compounded continuously, then A(t)=Pert. Here P=$500 and r=0.055. Therefore, A(t)=500e0.055t.
b. After 10 years, the amount of money in the account is
A(10)=500e0.055⋅10=500e0.55≈$866.63.
After 20 years, the amount of money in the account is
A(20)=500e0.055⋅20=500e1.1≈$1,502.08.
If $750 is invested in an account at an annual interest rate of 4%, compounded continuously, find a formula for the amount of money in the account after t years. Find the amount of money after 30 years.
- Hint
-
A(t)=Pert
- Answer
-
A(t)=750e0.04t. After 30 years, there will be approximately $2,490.09.
Logarithmic Functions
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 the pH scale in chemistry or decibels in sound levels.
The exponential function f(x)=bx is one-to-one, with domain (−∞,∞) and range (0,∞). Therefore, it has an inverse function, called the logarithmic function with base b. For any b>0,b≠1, the logarithmic function with base b, denoted logb, has domain (0,∞) and range (−∞,∞),and satisfies
logb(x)=y
if and only if by=x.
For example,
log2(8)=3
since 23=8,
log10(1100)=−2
since 10−2=1102=1100,
logb(1)=0
since b0=1 for any base b>0.
Furthermore, since y=logb(x) and y=bx are inverse functions,
logb(bx)=x
and
blogb(x)=x.
The most commonly used logarithmic function is the function loge. Since this function uses natural e as its base, it is called the natural logarithm. Here we use the notation ln(x) or lnx to mean loge(x). For example,
ln(e)=loge(e)=1ln(e3)=loge(e3)=3ln(1)=loge(1)=0.
Since the functions f(x)=ex and g(x)=ln(x) are inverses of each other,
ln(ex)=x and elnx=x,
and their graphs are symmetric about the line y=x (Figure 1.5.4).

In general, for any base b>0, b≠1, the function g(x)=logb(x) is symmetric about the line y=x with the function f(x)=bx. Using this fact and the graphs of the exponential functions, we graph functions logb for several values of b>1 ( Figure 1.5.5).

Before solving some equations involving exponential and logarithmic functions, let’s review the basic properties of logarithms.
If a,b,c>0,b≠1, and r is any real number, then
- Product property
logb(ac)=logb(a)+logb(c)
- Quotient property
logb(ac)=logb(a)−logb(c)
- Power property
logb(ar)=rlogb(a)
Solve each of the following equations for x.
- 5x=2
- ex+6e−x=5
Solution
a. Applying the natural logarithm function to both sides of the equation, we have
ln5x=ln2.
Using the power property of logarithms,
xln5=ln2.
Therefore,
x=ln2ln5.
b. Multiplying both sides of the equation by ex,we arrive at the equation
e2x+6=5ex.
Rewriting this equation as
e2x−5ex+6=0,
we can then rewrite it as a quadratic equation in ex:
(ex)2−5(ex)+6=0.
Now we can solve the quadratic equation. Factoring this equation, we obtain
(ex−3)(ex−2)=0.
Therefore, the solutions satisfy ex=3 and ex=2. Taking the natural logarithm of both sides gives us the solutions x=ln3,ln2.
Solve
e2x/(3+e2x)=1/2.
- Hint
-
First solve the equation for e2x
- Answer
-
x=ln32.
Solve each of the following equations for x.
- ln(1x)=4
- log10√x+log10x=2
- ln(2x)−3ln(x2)=0
Solution
a. By the definition of the natural logarithm function,
ln(1x)=4
- if and only if e4=1x.
Therefore, the solution is x=1/e4.
b. Using the product (Equation ???) and power (Equation ???) properties of logarithmic functions, rewrite the left-hand side of the equation as
log10√x+log10x=log10x√x=log10x3/2=32log10x.
Therefore, the equation can be rewritten as
32log10x=2
or
log10x=43.
The solution is x=104/3=103√10.
c. Using the power property (Equation ???) of logarithmic functions, we can rewrite the equation as ln(2x)−ln(x6)=0.
Using the quotient property (Equation ???), this becomes
ln(2x5)=0
Therefore, 2/x5=1, which implies x=5√2. We should then check for any extraneous solutions.
Solve ln(x3)−4ln(x)=1.
- Hint
-
First use the power property, then use the product property of logarithms.
- Answer
-
x=1e
When evaluating a logarithmic function with a calculator, you may have noticed that the only options are log10 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. 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, and a≠1,b≠1.
1. ax=bxlogba for any real number x.
If b=e, this equation reduces to ax=exlogea=exlna.
2. logax=logbxlogba for any real number x>0.
If b=e, this equation reduces to logax=lnxlna.
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, logb(ax)=xlogba. Therefore,
blogb(ax)=bxlogba.
In addition, we know that bx and logb(x) are inverse functions. Therefore,
blogb(ax)=ax.
Combining these last two equalities, we conclude that ax=bxlogba.
To prove the second property, we show that
(logba)⋅(logax)=logbx.
Let u=logba,v=logax, and w=logbx. We will show that u⋅v=w. By the definition of logarithmic functions, we know that bu=a,av=x, and bw=x. From the previous equations, we see that
buv=(bu)v=av=x=bw.
Therefore, buv=bw. Since exponential functions are one-to-one, we can conclude that u⋅v=w.
◻
Use a calculating utility to evaluate log37 with the change-of-base formula presented earlier.
Solution
Use the second equation with a=3 and b=e: log37=ln7ln3≈1.77124.
Use the change-of-base formula and a calculating utility to evaluate log46.
- Hint
-
Use the change of base to rewrite this expression in terms of expressions involving the natural logarithm function.
- Answer
-
log46=ln6ln4≈1.29248
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 R1 on the Richter scale and a second earthquake with magnitude R2 on the Richter scale. Suppose R1>R2, which means the earthquake of magnitude R1 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 A1 is the amplitude measured for the first earthquake and A2 is the amplitude measured for the second earthquake, then the amplitudes and magnitudes of the two earthquakes satisfy the following equation:
R1−R2=log10(A1A2).
Consider an earthquake that measures 8 on the Richter scale and an earthquake that measures 7 on the Richter scale. Then,
8−7=log10(A1A2).
Therefore,
log10(A1A2)=1,
which implies A1/A2=10 or A1=10A2. Since A1 is 10 times the size of A2, 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
log10(A1A2)=8−6=2.
Therefore, A1=100A2.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?
Solution
To compare the Japan and Haiti earthquakes, we can use an equation presented earlier:
9−7.3=log10(A1A2).
Therefore, A1/A2=101.7, and we conclude that the earthquake in Japan was approximately 50 times more intense than the earthquake in Haiti.
Compare the relative severity of a magnitude 8.4 earthquake with a magnitude 7.4 earthquake.
- Hint
-
R1−R2=log10(A1/A2).
- Answer
-
The magnitude 8.4 earthquake is roughly 10 times as severe as the magnitude 7.4 earthquake.
Hyperbolic Functions
The hyperbolic functions are defined in terms of certain combinations of ex and e−x. These functions arise naturally in various engineering and physics applications, including the study of water waves and vibrations of elastic membranes. Another common use for a hyperbolic function is the representation of a hanging chain or cable, also known as a catenary (Figure 1.5.7). If we introduce a coordinate system so that the low point of the chain lies along the y-axis, we can describe the height of the chain in terms of a hyperbolic function. First, we define the hyperbolic functions.

Hyperbolic cosine
coshx=ex+e−x2
Hyperbolic sine
sinhx=ex−e−x2
Hyperbolic tangent
tanhx=sinhxcoshx=ex−e−xex+e−x
Hyperbolic cosecant
cschx=1sinhx=2ex−e−x
Hyperbolic secant
sechx=1coshx=2ex+e−x
Hyperbolic cotangent
cothx=coshxsinhx=ex+e−xex−e−x
The name cosh rhymes with “gosh,” whereas the name sinh is pronounced “cinch.” Tanh,sech,csch, and coth are pronounced “tanch,” “seech,” “coseech,” and “cotanch,” respectively.
Using the definition of cosh(x) and principles of physics, it can be shown that the height of a hanging chain, such as the one in Figure 1.5.8, can be described by the function h(x)=acosh(x/a)+c for certain constants a and c.
But why are these functions called hyperbolic functions? To answer this question, consider the quantity cosh2t−sinh2t. Using the definition of cosh and sinh, we see that
cosh2t−sinh2t=e2t+2+e−2t4−e2t−2+e−2t4=1.
This identity is the analog of the trigonometric identity cos2t+sin2t=1. Here, given a value t, the point (x,y)=(cosht,sinht) lies on the unit hyperbola x2−y2=1 (Figure 1.5.8).

Graphs of Hyperbolic Functions
To graph coshx and sinhx, we make use of the fact that both functions approach (1/2)ex as x→∞, since e−x→0 as x→∞. As x→−∞,coshx approaches 1/2e−x, whereas sinhx approaches −1/2e−x. Therefore, using the graphs of 1/2ex,1/2e−x, and −1/2e−x as guides, we graph coshx and sinhx. To graph tanhx, we use the fact that tanh(0)=0, −1<tanh(x)<1 for all x, tanhx→1 as x→∞, and tanhx→−1 as x→−∞. The graphs of the other three hyperbolic functions can be sketched using the graphs of coshx, sinhx, and tanhx (Figure 1.5.9).

Identities Involving Hyperbolic Functions
The identity cosh2t−sinh2t=1, shown in Figure 1.5.8, is one of several identities involving the hyperbolic functions, some of which are listed next. The first four properties follow easily from the definitions of hyperbolic sine and hyperbolic cosine. Except for some differences in signs, most of these properties are analogous to identities for trigonometric functions.
- cosh(−x)=coshx
- sinh(−x)=−sinhx
- coshx+sinhx=ex
- coshx−sinhx=e−x
- cosh2x−sinh2x=1
- 1−tanh2x=sech2x
- coth2x−1=csch2x
- sinh(x±y)=sinhxcoshy±coshxsinhy
- cosh(x±y)=coshxcoshy±sinhxsinhy
- Simplify sinh(5lnx).
- If sinhx=3/4, find the values of the remaining five hyperbolic functions.
Solution:
a. Using the definition of the sinh function, we write
sinh(5lnx)=e5lnx−e−5lnx2=eln(x5)−eln(x−5)2=x5−x−52.
b. Using the identity cosh2x−sinh2x=1,we see that
cosh2x=1+(34)2=2516.
Since coshx≥1 for all x, we must have coshx=5/4. Then, using the definitions for the other hyperbolic functions, we conclude that tanhx=3/5,cschx=4/3,sechx=4/5, and cothx=5/3.
Simplify cosh(2lnx).
- Hint
-
Use the definition of the cosh function and the power property of logarithm functions.
- Answer
-
(x2+x−2)/2
Inverse Hyperbolic Functions
From the graphs of the hyperbolic functions, we see that all of them are one-to-one except coshx and sechx. If we restrict the domains of these two functions to the interval [0,∞), then all the hyperbolic functions are one-to-one, and we can define the inverse hyperbolic functions. Since the hyperbolic functions themselves involve exponential functions, the inverse hyperbolic functions involve logarithmic functions.
sinh−1x=arcsinhx=ln(x+√x2+1)cosh−1x=arccoshx=ln(x+√x2−1)tanh−1x=arctanhx=12ln(1+x1−x)coth−1x=arccotx=12ln(x+1x−1)sech−1x=arcsechx=ln(1+√1−x2x)csch−1x=arccschx=ln(1x+√1+x2|x|)
Let’s look at how to derive the first equation. The others follow similarly. Suppose y=sinh−1x. Then, x=sinhy and, by the definition of the hyperbolic sine function, x=ey−e−y2. Therefore,
ey−2x−e−y=0.
Multiplying this equation by ey, we obtain
e2y−2xey−1=0.
This can be solved like a quadratic equation, with the solution
ey=2x±√4x2+42=x±√x2+1.
Since ey>0,the only solution is the one with the positive sign. Applying the natural logarithm to both sides of the equation, we conclude that
y=ln(x+√x2+1).
Evaluate each of the following expressions.
sinh−1(2)
tanh−1(1/4)
Solution:
sinh−1(2)=ln(2+√22+1)=ln(2+√5)≈1.4436
tanh−1(1/4)=12ln(1+1/41−1/4)=12ln(5/43/4)=12ln(53)≈0.2554
Evaluate tanh−1(1/2).
- Hint
-
Use the definition of tanh−1x and simplify.
- Answer
-
12ln(3)≈0.5493.
Key Concepts
- The exponential function y=bx is increasing if b>1 and decreasing if 0<b<1. Its domain is (−∞,∞) and its range is (0,∞).
- The logarithmic function y=logb(x) is the inverse of y=bx. Its domain is (0,∞) and its range is (−∞,∞).
- The natural exponential function is y=ex and the natural logarithmic function is y=lnx=logex.
- Given an exponential function or logarithmic function in base a, we can make a change of base to convert this function to any base b>0, b≠1. We typically convert to base e.
- The hyperbolic functions involve combinations of the exponential functions ex and e−x. As a result, the inverse hyperbolic functions involve the natural logarithm.
Glossary
- base
- the number b in the exponential function f(x)=bx and the logarithmic function f(x)=logbx
- exponent
- the value x in the expression bx
- hyperbolic functions
- the functions denoted sinh,cosh,tanh,csch,sech, and coth, which involve certain combinations of ex and e−x
- inverse hyperbolic functions
- the inverses of the hyperbolic functions where cosh and sech are restricted to the domain [0,∞);each of these functions can be expressed in terms of a composition of the natural logarithm function and an algebraic function
- natural exponential function
- the function f(x)=ex
- natural logarithm
- the function lnx=logex
- number e
- as m gets larger, the quantity (1+(1/m)m gets closer to some real number; we define that real number to be e; the value of e is approximately 2.718282