EXPONENTIAL AND LOGARITHMIC FUNCTIONS

Some bacteria reproduce very quickly, as you may have discovered if you have ever had an infected cut or strep throat. Under the right circumstances, the number of bacteria in certain cultures will double in as little as an hour (or even less). In this section, we discuss some functions that can be used to model such rapid growth.

For instance, suppose that initially there are 100 bacteria at a given site, and the pop-ulation doubles every hour. Call the poppop-ulation function P(t), where t represents time (in hours) and start the clock running at time t= 0. The initial population of 100 means that P(0)= 100. After 1 hour, the population will double to 200, so that P(1) = 200. After another hour, the population will have doubled again to 400, making P(2)= 400. After a third hour, the population will have doubled again to 800, making P(3)= 800 and so on.

Let’s compute the population after 10 hours. (Try guessing this now. Most people guess poorly on this type of problem.) You could calculate the population at 4 hours, 5 hours, and so on, or you could use the following shortcut. To find P(1), you double the initial population, so that P(1)= 2 · 100. To find P(2), you double the population at time t = 1, so that P(2)= 2 · 2 · 100 = 2²· 100. Similarly, P(3) = 2³· 100. This pattern leads us to

P(10)= 2¹⁰· 100 = 102,400.

Depending on the organism, this is now a population that could cause some trouble!

The pattern just discovered suggests that the population can be modeled by the function P(t)= 2^t· 100.

We call P(t) an exponential function because the variable t is in the exponent. There is a subtle question here: what is the domain of this function? We have so far used only integer values of t, but for what other values of t does P(t) make sense? Certainly, rational powers

make sense, as in P(1/2) = 2^1/2· 100, where you recognize that 2^1/2=√

2. This says that the number of bacteria in the culture after a half hour is

P(1/2) = 2^1/2· 100 =√

2· 100 ≈ 141.

In this context, the function is not intended to give the exact number of bacteria at a given time, but rather, to predict an approximate number.

In any case, it’s a simple matter to interpret fractional powers as roots. For instance, x^1/2=√

x, x^1/3=√³

x, x^1/4 =√⁴

x, and so on. Further, we can write x^2/3=√³

x²=√₃ x2

, x^7/4=√⁴

x⁷=√₄ x7

, x^3.1= x^31/10 = √¹⁰

x³¹

and so on. But, what about irrational powers? These are somewhat harder to define, but they work exactly the way you would want them to. For instance, sinceπ is between 3.14 and 3.15, 2^πis between 2^3.14and 2^3.15. In this way, we define 2^xfor x irrational to fill in the gaps in the graph of y= 2^xfor x rational. That is, if x is irrational and a< x < b, for rational numbers a and b, then 2^a< 2^x < 2^b. This is the logic behind the definition of irrational powers.

If for some reason, you want to find the population afterπ hours, you can use your calculator or computer to obtain the approximate population:

P(π) = 2^π · 100 ≈ 882.

For your convenience, we summarize the usual rules of exponents next.

RULES OF EXPONENTS r For any integers m and n,

x^m/n=√ⁿ

x^m=√_n xm

. r For any real number p,

x^−p= 1 x^p. r For any real numbers p and q,

(x^p)^q = x^p·q. r For any real numbers p and q,

x^p· x^q = x^p+q.

Throughout your calculus course, you will need to be able to quickly convert back and forth between exponential form and fractional or root form.

EXAMPLE 5.1 Converting Expressions to Exponential Form Convert each to exponential form: (a) 3√

x⁵, (b) 5

√3

x, (c) 3x² 2√

x, and (d) (2^x· 2^3+x)². Solution For (a), simply leave the 3 alone and convert the power:

3√

x⁵= 3x^5/2.

For (b), use a negative exponent to write x in the numerator:

5

√3

x = 5x^−1/3.

For (c), first separate the constants from the variables and then simplify:

3x² 2√

x = 3 2

x² x^1/2 = 3

2x^2−1/2= 3 2x^3/2. For (d), first work inside the parentheses and then square:

(2^x· 2^3+x)² = (2^x+3+x)²= (2^2x+3)²= 2^4x+6. 

The function in part (d) of example 5.1 is called an exponential function with a base of 2.

DEFINITION 5.1

For any positive constant b= 1, the function f (x) = b^xis called an exponential function. Here, b is called the base and x is the exponent.

Be careful to distinguish between algebraic functions like f (x)= x³and g(x)= x^2/3 and exponential functions. Confusing these types of functions is a very common error.

Notice that for exponential functions like h(x)= 2^x, the variable is in the exponent (hence the name), instead of in the base. Also, notice that the domain of an exponential function is the entire real line, (−∞, ∞), while the range is the open interval (0, ∞).

While any positive real number can be used as a base for an exponential function, three bases are the most commonly used in practice. Base 2 arises naturally when analyzing processes that double at regular intervals (such as the bacteria at the beginning of this section). Also, most computers perform their calculations using base 2 arithmetic. Our standard counting system is base 10, so this base is commonly used. However, far and away the most useful base is the irrational number e. Likeπ, the number e has a surprising tendency to occur in important calculations. We define e by the following:

e= lim_n

→∞

 1+1

n

n

. (5.1)

Note that equation (5.1) has at least two serious shortcomings. First, we have not yet said what the notation lim

n→∞means. (In fact, we won’t define this until Chapter 1.) Second, it’s unclear why anyone would ever define a number in such a strange way. We will not be in a position to answer the second question until Chapter 6 (but the answer is worth the wait).

It suffices for the moment to say that equation (5.1) means that e can be approximated by calculating values of (1+ 1/n)ⁿ for large values of n and that the larger the value of n, the closer the approximation will be to the actual value of e. In particular, if you look at the sequence of numbers (1+ 1/2)², (1 + 1/3)³, (1 + 1/4)⁴ and so on, they will get progressively closer and closer to (i.e., home-in on) the irrational number e (named by the famous mathematician Leonhard Euler).

To get an idea of the value of e, we compute several of these numbers:

 1+ 1

10

10

= 2.5937 . . . ,

 1+ 1

1000

₁₀₀₀

= 2.7169 . . . ,



1+ 1

10,000

_10,000

= 2.7181 . . .

and so on. You can compute enough of these values to convince yourself that the first few digits of the decimal representation of e (e≈ 2.718281828459 . . .) are correct.

EXAMPLE 5.2 Computing Values of Exponentials Approximate e⁴, e^−1/5and e⁰.

Solution From a calculator, we find that

e⁴= e · e · e · e ≈ 54.598.

From the usual rules of exponents, e^−1/5= 1

e^1/5 = 1

√5

e≈ 0.81873.

(On a calculator, it is convenient to replace−1/5 with −0.2.) Finally, e⁰= 1.  The graphs of the exponential functions summarize many of their important properties.

EXAMPLE 5.3 Sketching Graphs of Exponentials

Sketch the graphs of the exponential functions y= 2^x, y = e^x, y = e^2x, y = e^x/2, y= (1/2)^xand y= e^−x.

Solution Using a calculator or computer, you should get graphs similar to those that follow.

y

x 4

2 2

4 10 20 30

FIGURE 0.64a y= 2^x.

Notice that each of the graphs in Figures 0.64a, 0.64b, 0.65a and 0.65b starts very near the x-axis (reading left to right), passes through the point (0, 1) and then rises steeply.

This is true for all exponentials with base greater than 1 and with a positive coefficient in the exponent. Note that the larger the base (e> 2) or the larger the coefficient in the exponent (2> 1 > 1/2), the more quickly the graph rises to the right (and drops to the left). Note that the graphs in Figures 0.66a and 0.66b (on the following page) are the mirror images in the y-axis of Figures 0.64a and 0.64b, respectively. The graphs rise as you move to the left and drop toward the x-axis as you move to the right. It’s worth noting that by the rules of exponents, (1/2)^x = 2^−x and (1/e)^x= e^−x.

y

x 4

2 2

4 10 20 30

FIGURE 0.64b y= e^x.

y

x 4

2 2

4 10 20 30

y

x 4

2 2

4 10 20 30

FIGURE 0.65a y= e^2x.

FIGURE 0.65b y= e^x/2.

y

x 4

2 2

4 10 20 30

y

x 4

2 2

4 10 20 30

FIGURE 0.66a y= (1/2)^x.

FIGURE 0.66b y= e^−x.



Notice in Figures 0.64–0.66 that each exponential function is one-to-one and hence, has an inverse function. We define the logarithmic functions to be inverses of the exponential functions.

DEFINITION 5.2

For any positive number b= 1, the logarithm function with base b, written logbx, is defined by

y= logbx if and only if x = b^y.

That is, the logarithm log_bx gives the exponent to which you must raise the base b to get the given number x. For example,

log₁₀10= 1 (since 10¹= 10), log₁₀100= 2 (since 10²= 100), log₁₀1000= 3 (since 10³= 1000)

and so on. The value of log₁₀45 is less clear than the preceding three values, but the idea is the same: you need to find the number y such that 10^y= 45. The answer is between 1 and 2 (why?), but to be more precise, you will need to employ trial and error. (Of course, you can always use your calculator or computer to compute an approximate value, but that won’t help you to understand what logarithms are all about.) You should get log₁₀45≈ 1.6532.

Later in the course, we introduce you to a powerful method for accurately approximating the values of such functions.

Observe from Definition 5.2 that for any base b> 0 (b = 1), if y = logbx, then x = b^y> 0. That is, the domain of f (x) = logbx is the interval (0,∞). Likewise, the range of

f is the entire real line, (−∞, ∞).

As with exponential functions, the most useful bases turn out to be 2, 10 and e. We usually abbreviate log₁₀x by log x. Similarly, logex is usually abbreviated ln x (for natural logarithm).

EXAMPLE 5.4 Evaluating Logarithms

Without using your calculator, determine log(1/10), log(0.001), ln e and ln e³.

Solution Since 1/10 = 10⁻¹, log(1/10) = −1. Similarly, since 0.001 = 10⁻³, we have that log(0.001) = −3. Since ln e = logee¹, ln e = 1. Similarly, ln e³= 3. 

We want to emphasize the inverse relationship defined by Definition 5.2. That is, b^x and log_bx are inverse functions for any b> 0.

In particular, for the base e, we have

e^{ln x} = x for any x > 0 and ln (e^x)= x for any x. (5.2)

We demonstrate this as follows. Let

y= ln x = logex.

By Definition 5.2, we have that

x= e^y= e^{ln x}.

We can use this relationship between natural logarithms and exponentials to solve equations involving logarithms and exponentials, as in examples 5.5 and 5.6.

EXAMPLE 5.5 Solving a Logarithmic Equation Solve the equation ln (x+ 5) = 3 for x.

Solution Taking the exponential of both sides of the equation and writing things back-ward (for convenience), we have

e³= e^{ln (x+5)}= x + 5, from (5.2). Subtracting 5 from both sides gives us

e³− 5 = x. 

EXAMPLE 5.6 Solving an Exponential Equation Solve the equation e^x+4= 7 for x.

Solution Taking the natural logarithm of both sides and writing things backward (for simplicity), we have from (5.2) that

ln 7= ln (e^x+4)= x + 4.

Subtracting 4 from both sides yields

ln 7− 4 = x. 

y

x

1 2 3 4 5

3

2

1 1 2

FIGURE 0.67a y= log x.

y

x

1 2 3 4 5

3

2

1 1 2

FIGURE 0.67b y= ln x.

As always, graphs provide excellent visual summaries of the important properties of a function.

EXAMPLE 5.7 Sketching Graphs of Logarithms

Sketch graphs of y= log x and y = ln x and briefly discuss the properties of each.

Solution From a calculator or computer, you should obtain graphs resembling those in Figures 0.67a and 0.67b. Notice that both graphs appear to have a vertical asymptote at x= 0 (why would that be?), cross the x-axis at x = 1 and very gradually increase as x increases. Neither graph has any points to the left of the y-axis, since log x and ln x are defined only for x > 0. The two graphs are very similar, although not identical. 

The properties just described graphically are summarized in Theorem 5.1.

THEOREM 5.1

For any positive base b= 1, (i) log_bx is defined only for x> 0, (ii) log_b1= 0 and

(iii) if b> 1, then logbx< 0 for 0 < x < 1 and logbx> 0 for x > 1.

PROOF

(i) Note that since b> 0, b^y> 0 for any y. So, if logbx = y, then x = b^y > 0.

(ii) Since b⁰= 1 for any number b = 0, logb1= 0 (i.e., the exponent to which you raise the base b to get the number 1 is 0).

(iii) We leave this as an exercise.

All logarithms share a set of defining properties, as stated in Theorem 5.2.

THEOREM 5.2

For any positive base b= 1 and positive numbers x and y, we have (i) log_b(x y)= logbx+ logby,

(ii) log_b(x/y) = logbx− logby (iii) log_b(x^y)= y logbx.

As with most algebraic rules, each of these properties can dramatically simplify calcu-lations when they apply.

EXAMPLE 5.8 Simplifying Logarithmic Expressions

Write each as a single logarithm: (a) log₂27^x− log₂3^x and (b) ln 8− 3 ln (1/2).

Solution First, note that there is more than one order in which to work each problem.

For part (a), we have 27= 3³and so, 27^x= (3³)^x= 3^3x. This gives us log₂27^x− log23^x= log23^3x− log23^x

= 3x log23− x log23= 2x log23= log23^2x. For part (b), note that 8= 2³and 1/2 = 2⁻¹. Then,

ln 8− 3 ln (1/2) = 3 ln 2 − 3(− ln 2)

= 3 ln 2 + 3 ln 2 = 6 ln 2 = ln 2⁶= ln 64.

In some circumstances, it is beneficial to use the rules of logarithms to expand a given expression, as in the following example.

EXAMPLE 5.9 Expanding a Logarithmic Expression Use the rules of logarithms to expand the expression ln

x³y⁴ z⁵

 .

Solution From Theorem 5.2, we have that ln

x³y⁴ z⁵



= ln (x³y⁴)− ln (z⁵)= ln (x³)+ ln (y⁴)− ln (z⁵)

= 3 ln x + 4 ln y − 5 ln z. 

Using the rules of exponents and logarithms, notice that we can rewrite any exponential as an exponential with base e, as follows. For any base a > 0, we have

a^x = e^{ln (ax)}= e^{x ln a}. (5.3)

This follows from Theorem 5.2 (iii) and the fact that e^{ln y} = y, for all y > 0.

EXAMPLE 5.10 Rewriting Exponentials as Exponentials with Base e Rewrite the exponentials 2^x, 5^x and (2/5)^xas exponentials with base e.

Solution From (5.3), we have

2^x= e^{ln (2x)}= e^{x ln 2}, 5^x= e^{ln (5x)}= e^{x ln 5}

and 

2 5

x

= e^ln[(2/5)x] = e^{x ln (2/5)}.



Just as we were able to use the relationship between the natural logarithm and exponen-tials to rewrite an exponential with any positive base in terms of an exponential with base e, we can use these same properties to rewrite any logarithm in terms of natural logarithms, as follows. For any positive base b (b= 1), we will show that

logbx= ln x

ln b. (5.4)

Let y = logbx. Then by Definition 5.2, we have that x= b^y. Taking the natural logarithm of both sides of this equation, we get by Theorem 5.2 (iii) that

ln x= ln (b^y)= y ln b.

Dividing both sides by ln b (since b= 1 and ln b = 0) gives us y= ln x

ln b, establishing (5.4).

One use you will find for equation (5.4) is for computing logarithms with bases other than e or 10. More than likely, your calculator has keys only for ln x and log x. We illustrate this idea in example 5.11.

EXAMPLE 5.11 Approximating the Value of Logarithms Approximate the value of log₇12.

Solution From (5.4), we have

log₇12=ln 12

ln 7 ≈ 1.2769894.



Hyperbolic Functions

There are two special combinations of exponential functions, called the hyperbolic sine and hyperbolic cosine functions, that have important applications. For instance, the Gateway Arch in Saint Louis was built in the shape of a hyperbolic cosine graph. (See the photograph in the margin.) The hyperbolic sine function is denoted by sinh (x) and the hyperbolic cosine function is denoted by cosh (x). These are defined to be

sinh x = e^x− e^−x

2 and cosh x =e^x+ e^−x

2 .

Graphs of these functions are shown in Figures 0.68a and 0.68b. The hyperbolic functions (including the hyperbolic tangent, tanh x, defined in the obvious way) are often convenient to use when solving equations. For now, we verify several basic properties that the hyperbolic functions satisfy in parallel with their trigonometric counterparts.

Saint Louis Gateway Arch.

4

2 2

4

10 10

5 5 y

x

4

2 2

4

10 10

5 5 y

x

FIGURE 0.68a y= sinh x.

FIGURE 0.68b y= cosh x.

EXAMPLE 5.12 Computing Values of Hyperbolic Functions

Compute f (0), f (1) and f (−1) and determine how f (x) and f (−x) compare for each function: (a) f (x)= sinh x and (b) f (x) = cosh x.

Solution For part (a), we have sinh 0=e⁰− e⁻⁰

2 = 1− 1

2 = 0. Note that this means that sinh 0= sin 0 = 0. Also, we have sinh 1 = e¹− e⁻¹

2 ≈ 1.18, while sinh (−1) = e⁻¹− e¹

2 ≈ −1.18. Notice that sinh (−1) = − sinh 1. In fact, for any x, sinh (−x) = e^−x− e^x

2 =−(e^x− e^−x)

2 = − sinh x.

[Recall that the same rule holds for the corresponding trigonometric function:

sin (−x) = −sin x.] For part (b), we have cosh 0 = e⁰+ e⁻⁰

2 = 1+ 1

2 = 1. Note that this means that cosh 0= cos 0 = 1. Also, we have cosh 1 =e¹+ e⁻¹

2 = ≈ 1.54, while cosh (−1) = e⁻¹+ e¹

2 ≈ 1.54. Notice that cosh (−1) = cosh 1. In fact, for any x, cosh (−x) = e^−x+ e^x

2 =e^x+ e^−x

2 = cosh x.

[Again, the same rule holds for the corresponding trigonometric function: cos (−x) = cos x.] 

Fitting a Curve to Data

You are familiar with the idea that two points determine a straight line. As we see in example 5.13, two points will also determine an exponential function.

EXAMPLE 5.13 Matching Data to an Exponential Curve

Find the exponential function of the form f (x)= ae^bx that passes through the points (0, 5) and (3, 9).

Solution Notice that you can’t simply solve this problem by inspection. That is, you can’t just read off appropriate values for a and b. Instead, we solve for a and b, using the properties of logarithms and exponentials. First, for the graph to pass through the point (0, 5), this means that

5= f (0) = ae^{b· 0}= a,

so that a= 5. Next, for the graph to pass through the point (3, 9), we must have 9= f (3) = ae^3b = 5e^3b.

To solve for the b in the exponent, we divide both sides of the last equation by 5 and take the natural logarithm of both sides, which yields

ln

9 5



= ln e^3b= 3b ln e = 3b,

since ln e= 1. Finally, dividing by 3 gives us the value for b:

b=1 3ln

9 5

 .

Thus, f (x)= 5e⁽¹/3) ln (9/5)x. 

2 3

1 4 5 6 7 8

5 10 15 20 25 30 35

Number of decades since 1780

U.S. population (in millions)

FIGURE 0.69a

U.S. Population 1790–1860.

Consider the population of the United States from 1790 to 1860, found in the ac-companying table. A plot of these data points can be seen in Figure 0.69a (where the vertical scale represents the population in millions). This shows that the population was increasing, with larger and larger increases each decade. If you sketch an imaginary curve through these points, you will probably get the impression of a parabola or perhaps the right half of a cubic or exponential. And that’s the question: is this data best modeled by a quadratic function, a cubic function, a fourth-order polynomial, an exponential function or what?

Year U.S. Population 1790 3,929,214 1800 5,308,483 1810 7,239,881 1820 9,638,453 1830 12,866,020 1840 17,069,453 1850 23,191,876 1860 31,443,321

We can use the properties of logarithms from Theorem 5.2 to help determine whether a given set of data is modeled better by a polynomial or an exponential function, as follows.

Suppose that the data actually comes from an exponential, say y= ae^bx(i.e., the data points lie on the graph of this exponential). Then,

ln y= ln (ae^bx)= ln a + ln e^bx = ln a + bx.

If you draw a new graph, where the horizontal axis shows values of x and the vertical axis corresponds to values of ln y, then the graph will be the line ln y= bx + c (where the

constant c= ln a). On the other hand, suppose the data actually came from a polynomial.

If y= bxⁿ(for any n), then observe that

ln y= ln (bxⁿ)= ln b + ln xⁿ= ln b + n ln x.

In this case, a graph with horizontal and vertical axes corresponding to x and ln y, respec-tively, will look like the graph of a logarithm, ln y= n ln x + c. Such a semi-log graph (i.e., graphing ln y versus x) lets us distinguish the graph of an exponential from that of a polynomial: graphs of exponentials become straight lines, while graphs of polynomials (of degree≥ 1) become logarithmic curves. Scientists and engineers frequently use semi-log graphs to help them gain an understanding of physical phenomena represented by some collection of data.

EXAMPLE 5.14 Using a Semi-Log Graph to Identify a Type of Function

Determine if the population of the United States from 1790 to 1860 was increasing exponentially or as a polynomial.

2

1 3 4 5 6 7 8 1

2 3 4

Number of decades since 1780 Natural logarithm of U.S. population (millions)

FIGURE 0.69b

Semi-log plot of U.S. population.

Solution As already indicated, the trick is to draw a semi-log graph. That is, instead of plotting (1, 3.9) as the first data point, plot (1, ln 3.9) and so on. A semi-log plot of this data set is seen in Figure 0.69b. Although the points are not exactly colinear (how would you prove this?), the plot is very close to a straight line with ln y-intercept of 1 and slope 0.35. You should conclude that the population is well-modeled by an exponential function. (Keep in mind that here, as with most real problems, the data is somewhat imprecise and so, the points in the semi-log plot need not be perfectly colinear for you to conclude that the data is modeled quite well by an exponential.) The exponential model would be y= P(t) = ae^bt, where t represents the number of decades since 1780. Here, b is the slope and ln a is the ln y-intercept of the line in the semi-log graph. That is, b= 0.35 and ln a = 1 (why?), so that a = e. The population is then modeled by

P(t)= e · e^0.35tmillion. 

EXERCISES 0.5

WRITING EXERCISES

1. Starting from a single cell, a human being is formed by 50 gen-erations of cell division. Explain why after n divisions there are 2ⁿ cells. Guess how many cells will be present after 50 divisions, then compute 2⁵⁰. Briefly discuss how rapidly expo-nential functions increase.

2. Explain why the graphs of f (x)= 2^−x and g(x)=₁

2

x

are the same.

3. Compare f (x)= x² and g(x)= 2^x for x= ¹₂, x = 1, x= 2, x = 3 and x = 4. In general, which function is bigger for large values of x? For small values of x?

4. Compare f (x)= 2^x and g(x)= 3^x for x= −2, x = −¹₂,

x= ¹₂ and x= 2. In general, which function is bigger for neg-ative values of x? For positive values of x?

In exercises 1–6, convert each exponential expression into frac-tional or root form.

1. 2⁻³ 2. 4⁻² 3. 3^1/2

4. 6^2/5 5. 5^2/3 6. 4^−2/3

In exercises 7–12, convert each expression into exponential form.

7. 1

x² 8. √³

x² 9. 2

x³ 10. 4

x² 11. 1

2√

x 12. 3

2√ x³

In exercises 13–16, find the integer value of the given expression without using a calculator.

13. 4^3/2 14. 8^2/3 15.

√8

2^1/2 16. 2

(1/3)² In exercises 17–20, use a calculator or computer to estimate each value.

17. 2e^−1/2 18. 4e^−2/3 19. 12

e 20. 14

√e In exercises 21–30, sketch a graph of the given function.

21. f (x)= e^2x 22. f (x)= e^3x 23. f (x)= 2e^x/4 24. f (x)= e^−x2 25. f (x)= 3e^−2x 26. f (x)= 10e^−x/3 27. f (x)= ln 2x 28. f (x)= ln x² 29. f (x)= e^{2 ln x} 30. f (x)= e^−x/4sin x In exercises 31–40, solve the given equation for x.

31. e^2x= 2 32. e^4x= 3

33. e^x(x²− 1) = 0 34. xe^−2x+ 2e^−2x= 0

35. ln 2x= 4 36. 2 ln 3x= 1

37. 4 ln x= −8 38. x²ln x− 9 ln x = 0 39. e^{2 ln x}= 4 40. ln (e^2x)= 6

In exercises 41 and 42, use the definition of logarithm to deter-mine the value.

41. (a) log₃9 (b) log₄64 (c) log₃₂₇¹ 42. (a) log₄₁₆¹ (b) log₄2 (c) log₉3

In exercises 43 and 44, use equation (5.4) to approximate the value.

43. (a) log₃7 (b) log₄60 (c) log₃₂₄¹ 44. (a) log₄₁₀¹ (b) log₄3 (c) log₉8

In exercises 45–50, rewrite the expression as a single logarithm.

45. ln 3− ln 4 46. 2 ln 4− ln 3 47. ¹₂ln 4− ln 2 48. 3 ln 2− ln¹₂ 49. ln³₄+ 4 ln 2 50. ln 9− 2 ln 3

In exercises 51–54, find a function of the form f (x) ae^bxwith the given function values.

51. f (0)= 2, f (2) = 6 52. f (0)= 3, f (3) = 4 53. f (0)= 4, f (2) = 2 54. f (0)= 5, f (1) = 2 55. A fast-food restaurant gives every customer a game ticket. With

each ticket, the customer has a 1-in-10 chance of winning a free meal. If you go 10 times, estimate your chances of winning at least one free meal. The exact probability is 1−₉

10

10

. Com-pute this number and compare it to your guess.

56. In exercise 55, if you had 20 tickets with a 1-in-20 chance of

在文檔中 The Real Number System (頁 46-57)