4 Chapter 4

5.11     InverseFunctionsandLogarithmsIf we think of a function as a process, then we can seek another function which might reverse that process. As in real life, we will determine that some processes (like putting on socks and shoes) are reversible, while other processes (like cooking a steak) are not.

 

Learning Objectives: In this section, you will learn about the properties and characteris­ tics of logarithmic functions. Upon completion you will be able to:

 

•    Justify whether or not a function is one-to-one.

•    Identify a logarithmic function.

•    Convert equations between exponential and logarithmic form.

•    Apply the laws and properties of logarithms to expand, condense, and simplify loga- rithmic expressions.

•    Memorize the graph of the parent logarithmic function base a.

•    Determine the domain of a logarithmic function, using interval notation.

•    Solve equations involving exponential functions with different bases.

•    Solve equations involving logarithms.

•    Use exponential and logarithmic functions to model and solve real-world applications.

 

 

 

Defining an Inverse Function

Let’s begin by discussing a very basic function which is reversible, f(x) = 3x + 4. Thinking off  as a process, we start with an input, x, and apply two steps:

 

1.   Multiply by 3

2.   Add 4

 

To reverse this process, we seek a function, g, which will undo each of these steps and take the output from f, 3x + 4, and return the input, x. If we think of the real-world reversible two step process of first putting on socks, and then putting on shoes, to reverse the process,

we first take off the shoes, and then we take off the socks. In much the same way, the function

g should undo the second step off  first. That is, function g should

 

1.   Subtract 4

2.   Divide by 3

Following this procedure, we get g(x) = xt.

 

;

Let’s check to see if the function g reverses the process of f(x). If x = 5, then f(5) = 3(5) + 4 = 15 + 4 = 19. Taking the output 19 from f, we substitute it into g to get g(19) = 19 4 = \5 = 5, which is our original input to f.

 

3

To check that g reverses the process of f ( x) for all x in the domain of f, we take the generic output from  f,  f(x) = 3x + 4, and substitute into g.  That is g(f (x)) = f(3x + 4) = (3x+ 4)-4 = 3f = x, which is our original input to f. If we carefully examine the arithmetic as

we simplify g(f(x)), we actually see g first ‘undoing’ the addition of 4, and then ‘undoing’ the multiplication by 3.

Not only does g undo f, but f also undoes g. That is, if we take the output from g, g(x) =

x34, and substitute that into J, we get J(g(x)) = f (x34) = 3 (x34) + 4 = (x – 4) + 4 = x. Using the language of composition, the statements g(f(x)) = x and f(g(x)) = x can be written as (go f)(x) = x and (f o g)(x) = x, respectively.

 

Abstractly, we can visualize the relationship between f and g in Figure 5.8.2 below.

 

 

 

Figure 5.8.2: A visual representation of the functions f and g ‘undoing’ each others processes.

 

From the diagram above we see that g takes the outputs from f and returns them to their respective inputs, and conversely, f takes outputs from g and returns them to their respective inputs. We now have enough background to state the following definition.

 

Note: The notation 1-1 is an unfortunate choice, as you might think of this as 7. This is definitely not the case.

Recall  f(x) = 3x + 4 has as its inverse 1–1(x) = x;4, which is certainly different than

1                1

f(x)–        3x+4·

 

Besides using compositions, one way to determine if a function is invertible is to determine if the function is one-to-one.

Definition 5.25

 

Graphically, we detect one-to-one functions using the test below.

Theorem 11 The Horizontal Line Test

Theorem 12 Equivalent Conditions for lnvertibility

Example 1: Determine if the following functions are one-to-one, using the Horizontal Line Test.

(a)       g(x) = l x

(b)        h(x)=x2-2x+4

Solution:     (a) We begin by graphing g(x) = l x  in our calculator, as seen in Figure 5.8.3.

Figure 5.8.3: Calculator screenshot of g(x).

One can imagine any horizontal line drawn through the graph will only intersect the curve at most once. Therefore, the graph of g(x) passes the Horizontal Line Test, and g(x) is a one-to-one function. Thus, g(x) is also invertible.

(b) Again, we begin by graphing h(x) = x2 – 2x + 4 in our calculator. (See Figure 5.8.4.)

Figure 5.8.4: Calculator screenshot of h(x).

We see immediately from the graph that h(x) fails the Horizontal Line Test, as there are several horizontal lines which cross the graph more than once (and all we need is one horizontal line to cross more than once). Thus, h(x) is NOT a one-to-one function and, therefore, not invertible.

Example 2: Determine whether or not f(x) = 1 2x and g(x) = – x     + ½ are inverses of each other.

Solution: To determine if f and g are inverses, we must compute (go f)(x) and (f o g)(x). We first check that (go f)(x) = x for all x in the domain of f(x), which is (-oo, oo).

 

 

 

 

 

 

 

 

 

 

 

 

 

 

We now check that (f og)(x) = x for all x in the domain of g(x), which is also (-oo, oo).

(f o g)(x) = f(g(x))

 

=f(-}x+t)

1- 2 (- x + ½)

5

1 + 5x – 1

5

5x

5

=x

 

Thus, f(x) and g(x) are inverses of each other.

 

To check our result, we could graph f(x) and g(x) on the same set of axes and confirm their graphs are reflections of each other across the line y = x. Another way to verify our results is to make a table of values for each function and confirm that if (a, b) is on f (x), then (b, a) is on g(x).

 

Instead of graphing, we will first choose values for x to build Table 5.29 with f(x) values. Then, we will use the function values found in Table 5.29 as x-values to build Table 5.30 with g(x) values, and verify f(x) and g(x) are indeed inverses.

 

X	f(x)	(x, J(x))
-5	11/5	(-5, 11/5)
-3	7/5	(-3,7/5)
-1	3/5	(-1,3/5)
0	1/5	(0, 1/5)
2	-3/5	(2, -3/5)
4	-7/5	(4, -7/5)
6	-11/5	(6, -11/5)

Table 5.29: A chart of inputs, x, and their corresponding outputs, f(x).

 

X	g(x)	(x, g(x))
11/5	-10/2	(11/5, -5)
7/5	-6/2	(7/5, -3)
3/5	-2/2	(3/5, -1)
1/5	0	(1/5,0)
-3/5	4/2	(-3/5, 2)
-7/5	-8/2	(-7/5,4)
-11/5	12/2	(-11/5, 6)

Table 5.30: A chart of inputs, x, and their corresponding outputs, g(x).

 

The inputs come from the outputs in Table 5.29.

 

In the previous two examples, we used the Horizontal Line Test to show a function has an inverse and compositions to verify functions are inverses of each other. While it is common to next discuss techniques for computing an inverse function, 1-1(x), the topic is not a focus of this text. The authors will leave it to the reader to explore computing inverse functions outside of this text.

 

 

Defining Logarithmic FunctionsAccording to http://earthquake.usgs.gov/ earthquakes/ eqinthenews/, there were many earth­ quakes in 2010 and 2011. In 2010, a major earthquake struck Haiti, destroying or damaging over 285,000 homes. One year later, another, stronger earthquake devastated Honshu, Japan, destroying or damaging over 332,000 buildings. Even though both caused substantial dam­ age, the earthquake in 2011 was 100 times stronger than the earthquake in Haiti. How do we know? The magnitudes of earthquakes are measured on a scale known as the Richter Scale. The Richter Scale uses powers of 10 to classify the strength of an earthquake. The

 

Haitian earthquake registered a 7.0 on the Richter Scale, whereas the Japanese earthquake registered a 9.0.

 

In order to analyze the magnitude of earthquakes or compare the magnitudes of two different earthquakes, we need to be able to convert from exponential forms to a new form called logarithmic form. For example, suppose the amount of energy released from one earthquake was 500 times greater than the amount of energy released from another. To calculate the difference in magnitude involves the equation 10x = 500, where x represents the difference in magnitudes on the Richter Scale. How would we solve for x? We have not yet learned a method for solving an exponential equation of this type. None of the algebraic tools discussed

so far are sufficient to solve 10x = 500. We know that 102 = 100 and 103 = 1000, so it is

clear that x must be some value between 2 and 3, as y = f(x) = 10x is increasing. We can examine a graph, as in Figure 5.8.5, to better estimate the solution.

Figure 5.8.5: A graph of y = 10x and y = 500 on the same coordinate plane. Their intersection point is indicated.

Estimating from a graph, however, is imprecise. To identify an algebraic solution, we must introduce a new function. Observe that the graph of y = 10x in Figure 5.8.5 passes the Hori­ zontal Line Test. Thus, the exponential function y = 10x is one-to-one, and it has an inverse. Although not discussed in detail by the authors, it is the case with all invertible functions that you can simply interchange the x and y to find the function’s inverse. However, to represent the inverse function as a function of x, we must then solve the new equation for y. For y = 10x, after interchanging the variables we would get x = lQY and then need to solve for y.

 

To solve for y requires the removal of a variable from the exponent. A new function type,

called a logarithmic function (also called a ‘ log ‘), is necessary to perform this operation. So for y = 10x, the following work would be necessary to compute the inverse function, 1-1(x).

X = lQY

log10(x) = y

1-1(x) = log10(x)

 

Graphing this function, along with f(x) = 10x (as shown in Figure 5.8.6), we see the graph of 1-1(x) is, as it should be, the graph of f(x) reflected over the line y = x.

Figure 5.8.6: A graph of f(x) and its inverse, 1-1(x).

 

Definition 5.26

In general, the inverse of the exponential function f (x) logarithm function, and is denoted 1-1(x) = logb(x).

 

bx is called the base b

 

 

 

 

On a procedural level, as they are inverses, logs ‘undo’ exponentials. Consider the function f(x) = 2x_ When we evaluate f(3) = 23 = 8, the input 3 becomes the exponent on the base 2 to produce the real number 8. The function 1-1(x) = log2(x) then takes the number 8 as its input and returns the real number 3 (which was the exponent of 2) as its output. In

symbols, log2(8) = 3. More generally, log2(x) is the exponent you raise 2 to in order to get

x. Thus, log2(16) = 4, because 24 = 16.

 

The following defines a logarithmic function using exponent notation, instead of inverse notation.

Definition 5.28

 

 

Using Algebraic Properties of Logarithms

We introduced logarithmic functions as inverses of exponential functions.         Now, we will explore the algebraic properties of logarithms.

Converting Equations between Exponential and Logarithmic Forms

 

Example 3: Write the following logarithmic equations in exponential form.

 

(a)   log6(v16) = ½

(b)   log3(9) = 2

Solution: First, identify the values of b, x, and y. Then, write the equation in the form bY = x.

2′         y u

b = 6′ y = 1  and x =     ‘6

1

Therefore, the equation log6(v16) = ½ is equivalent to 6½ = v16, which is illustrated as

log6 ( v’6) = 2

means

 

(b) For log3(9) = 2:      b = 3, y = 2, and x = 9

Therefore, the equation log3 (9) = 2 is equivalent to 32 = 9, which is illustrated as log3(9) = 2 means  32 = 9

 

Try It 2

 

 

Example 4: Write the following exponential equations in logarithmic form.

(a)       23 = 8

(b)       52 = 25

(C ) 10-4 =

1

10000

Solution: First, identify the values of b, y, and x.  Then, write the equation in the form y =

logb(x).

(a)    For 23 = 8:      b = 2, y = 3, and x = 8

Therefore, the equation 23 = 8 is equivalent to log2(8) = 3.

(b)    For 52 = 25:      b = 5, y = 2, and x = 25

Therefore, the equation 52 = 25 is equivalent to log5 (25) = 2.

(c)    For 10-4 = 10600 :        b = 10, y = -4, and x = 106oo

Therefore, the equation 10-4 = 106oo is equivalent to log10Co6oo) = -4.

TryIt 3

 

Knowing the squares, cubes, and roots of numbers allows us to evaluate many logarithms mentally. For example, consider log2(16). We ask, “To what exponent must 2 be raised in order to get 16?” This is equivalent to log2(16) = y, which means 2Y = 16. Because we already know 16 = 24, then

2Y = 16

2Y = 24

y=4

 

Example 5: Evaluate log4(64), without a calculator.

Solution: First, we rewrite the logarithm, y = log4(64), in exponential form.

 

Then, we find a common base, and solve for y.

 

 

Try It 4

 

Suppose we wanted to evaluate log2(7). Proceeding in the manner just described gives us

The answer should be a little less than 3 given 7 < 8 and 8 = 23, but can we be more precise? Because 7 is not an integer power of 2, we can use the calculator to tell us a more accurate answer. The two logarithm buttons commonly found on calculators are the “LOG” and “LN” buttons, which correspond to the common and natural logs, respectively. As we have base 2 here, and not base 10 or base e, in order to use the calculator to give us a more accurate evaluation, we need the following theorem.

Theorem 13 Change of Base Formula

 

 

 

 

 

 

 

Returning to our evaluation of log2(7), we have

log(7)                           ln(7)

log2(7) =log(2)  or log2(7) =ln(2)

Using the calculator and the change of base formulas shown above, we can now obtain a closer approximation. (See Figures 5.8.7 and 5.8.8.)

 

 

Figure 5.8.7: A calculator approximation of log2(7) using the “LOG” button.

Figure 5.8.8: A calculator approximation of log2(7) using the “LN” button.

From the calculator we see that log2(7)      2.80735, which is indeed “a little less than 3.”

 

 

Condensing and Expanding Logarithms

As logarithmic and exponential functions ‘undo’ each other, logarithms have similar algebraic properties to exponentials. Some important properties of logarithms are given here.

 

Note: In the Product and Quotient Rules, the base of the logarithms must be the same in order to condense them into a single logarithm of the same base.

Note: Due to the fact that logb(b) = 1, it follows that ln(e) = 1 and log(lO) = 1. Here are some common mistakes to be aware of as you use these rules.

Example 6: For x, y, z > 0, use the properties of logarithms to write the following as a single logarithm.

log(x) – 2log(y) + log(z)

Solution: In the expression log( x) – 2 log(y) + log(z), we have both a difference and a sum of logarithms, all in the same base. However, before we use the Quotient Rule to combine log(x) – 2log(y), we need to address the coefficient, 2, of log(y). This can be handled using the Power Rule. We can then apply the Quotient and Product Rules, as we move from left to right. Putting this all together, we have

log(x) – 2log(y) + log(z) = log(x) – log (y2) + log(z)

= log (;2)   + log(z)

= log (;2•  z)

 

=log(;:)

 

Example 7: For x, y, z > 0, use the properties of logarithms to write the following as a single logarithm.

 

Solution: All three logarithms have the same base, 2, so we can use the given rules to con­ dense them into a single logarithm. Remember to use the Power Rule to address any coefficients, before using the Product and Quotient Rules.

 

log2 (x2) +    log2(x – 1) – 3log2 ((x + 3)2) = log2 (x2) + log2 ((x – 1)½) – log2 ( ((x + 3)2)3)

= log2 (x2) + log2 ((x – 1)½) – log2 ((x + 3)6)

= log2 ( x2(x – 1)½) – log2 ((x + 3)6)

It is sometimes necessary to use the facts that ln(e) = 1 and log(lO) = 1 when con­ densing or simplifying logarithms.

 

Example 8: Use the properties of logarithms to write the following as a single logarithm. Assume, when necessary, that all variables represent positive real numbers.

–

1- ln(x) –

2

Solution: First, we use the Power Rule to move the coefficient of the logarithm to the exponent of the variable.

– ln(x) – –1 = (–1)   ln(x) – -1

2                            2

= ln (x-1) – 1-

2

In order to simplify further ½ must be rewritten as a logarithm. Because ln(e) = 1,

 

= ln (x-1) –

= ln (x-1) –

1 (1)

1

2

2

(ln(e))

= ln (x-1) – ln ( e½)

 

Finally, the Quotient Rule gives us

 

1

= ln( :      )

=ln (x )

 

 

Example 9: Use the properties of logarithms to fully expand and simplify the following expression.

Assume, when necessary, that all arguments represent positive real numbers.

 

Solution: To expand log2 ( ), we use the Quotient Rule.

Due to the fact that 8 = 23 and log2(2) = 1, we can simplify and have

log2(8) – log2(x) = log2 (23) –  log2(x)

= 3 – log2(x)

= – log2(x) + 3

Example 10: Use the properties of logarithms to fully expand and simplify the following expression.

Assume, when necessary, that all arguments represent positive real numbers.

Solution: When expanding logarithms, before using the Quotient or Product Rules, we must first address any powers on the arguments of the logarithms.

Then, we can continue with the Quotient and Product Rules, as follows.

= 2[ln(3) – ln(ex)]

= 2[ln(3) – {ln(e) + ln(x)}]

To simplify this expression, distribute and remember ln(e) = 1.

= 2 ln(3) – 2[ln(e) + ln(x)]

= 2ln(3) – 2ln(e) – 2ln(x)]

= 2ln(3) – 2 – 2ln(x)

6

Example 11: Use the properties oflogarithms to fully expand and simplify log ( 64(;;’=\;1)).Assume, when necessary, that x > ½.

Solution:

 

64x3

4x + 1))                 3

log6(

(  x(

2

_ l)       = log6 (64x (4x + 1)) – log6(2x – 1)

= log6 (64×3) + log6(4x + 1) – log6(2x – 1)

= log6(64) + log6 (x3) + log6(4x + 1) – log6(2x – 1)

= log6 (26) + log6 (x3) + log6(4x + 1) – log6(2x – 1)

= 6log6(2) + 3log6(x) + log6(4x + 1) – log6(2x –  1)

 

 

Properties of Logarithmic Functions

Recall that the exponential function is defined as y = bx for any real number x and constant

b > 0, b #- 1, where

•    The domain of y is (-oo, oo).

•    The range of y is (0, oo).

We just learned that the logarithmic function y = logb(x) is the inverse of the exponential function y = bx. So, as the inverse function

•    The domain of y = logb(x) is the range of y =bx: (0, oo).

•    The range of y = logb(x) is the domain of y =bx: (-00,00).

The following summarizes this and other basic properties of logarithmic functions, all of which come from the fact that they are inverses of exponential functions and are exponents themselves.

Properties of Logarithmic Functions

 

We will add our final two parent functions to our list of parent functions, without including a table of values for each.

 

 

 

Table 5.31: Parent Logarithmic Functions

 

Example 12: Given h(x) = log5(x), state each of the following, if it exists.

 

(a)    Domain

(b)    Range

(c)   End behavior

(d)    x-intercept (s)

(e)    y-intercept (s)

 

Solution: We begin by recognizing b = 5, (b > 1).

 

(a)    The domain of h(x) is (0, oo).

(b)   The range of h(x) is (-00,00).

(c)   The end behavior of h(x) is As x—-+ 0 from the right, f(x) —-+ -oo. (There is a vertical asymptote at x = 0.) As x—-+ oo, f(x)—————————————- + oo.

(d)   h(x) has an x-intercept at (1, 0).

(e)    h ( x) has no y-intercept, as x = 0 is a vertical asymptote of h ( x).

 

 

 

 

Computing the Domain of a Logarithmic Function

Up until this point, restrictions on the domains of functions came from avoiding division by zero and keeping negative numbers from beneath even radicals. With the introduction of log­ arithms, we now have another restriction. Due to the fact that the domain of f(x) = logb(x) is (0, oo), the input, or argument, of the logarithm must be strictly positive.

Generally speaking, in order for f(x) = logb(g(x)) to be defined, g(x) must be defined AND

g(x) must be greater than zero.

 

Consider f(x) = log4(2x – 3). This function is defined for any values of x such that the argument, 2x – 3, is defined AND greater than zero. As g(x) is a linear polynomial, it is defined for all real numbers. We now set the argument greater than zero and solve for x to find any domain restrictions.

2x – 3 > 0

2x > 3

x> 2

So, the domain of f(x) is (too).

Example 13: State the domain of the following functions, using interval notation.

(a)     f(x) = 2 log(3 –  x) – 1

(b)   h(x) = ln C 1)

Solution:  (a) For f(x) to be defined, the argument, g(x) = 3- x, must be defined AND greater than zero. Given that g(x) = 3-x  is a linear polynomial, g(x) is defined for all real numbers. So, we set the argument greater than zero and solve for x, to find any domain restrictions.

3-x>O

-x > -3

x<3

Therefore, the domain of f(x) = 2 log(3 – x) – 1 is (-oo, 3).

(b) For h(x) to be defined, the argument, g(x) = x l’ must be defined AND greater than zero. Seeing as g(x) = x l  is a rational function, g(x) is defined for x =/- l. For x l  to be greater than 0 (positive), x – l must be positive, as 4 is always positive:

x-l>0

x>l

Therefore, the domain of h( x) is the overlapping intervals of x =/- l and x > l.

So, the domain of h(x) = ln C 1) is (1, oo).

 

 

Solving Equations Involving Exponentials

Recall the Common Base Property of Exponents (One-to-One Property of Exponents)

 

b5 = bT     if and only if      S = T

for all real numbers Sand T and b > 0, b =/- l. So when the bases are the same, the exponents are equal.

 

Similarly, there is a One-to-One Property for Logarithms.

Theorem 14 One-to-One Property of Logarithms

For all real numbers M > 0 and N > 0, logb(M) = logb(N) if and only if M = N.

 

Sometimes the terms of an equation involving exponentials cannot be rewritten with a com­ mon base. By the One-to-One Property of Logarithms, we solve such equations by applying the logarithm to each side. While you can use a logarithm with any base, the authors will choose to use the natural logarithm, unless the base of the exponential expression is 10. In this case, the authors will use the common logarithm. The authors make these choices in the text so that readers can quickly verify their answers using most calculators.

Example 14: Solve 5(x+2) = 4x for x. Leave all answers as exact values.

Solution: There is no easy way to rewrite all exponential expressions in this equation with the same base. Thus, we will use the One-to-One Property of Logarithms and apply the natural log to both sides of the equation.

5(x+2)= 4x

ln (5(x+2l) = ln(4x)

 

From here we will use the properties of logarithms to solve for x.

(x + 2) ln(5) = x ln(4)

x ln(5) + 2 ln(5) = x ln(4) xln(5) – xln(4) = -2ln(5)

x[ln(5) – ln(4)] = ln (5-2)

1

x  ln (:)      = ln ( 2 5)

ln U5)

X=  ln (i)

 

The solution to 5(x+2) = 4x ‘ x = iin((¾) ‘

is NOT the same as x = ln (l)   – ln (Q)    A

25                     4 •

quick check on your calculator will show the difference between the two.

Example 15: Solve 4e2x + 5 = 12 for x. Leave all answers as exact values.

Solution: We begin by moving and combining like terms so that we have a single term on each side of the equation.

4e2x + 5 = 12

4e2x = 7

Before applying the natural log of both sides, it is a best practice to divide both sides of the equation by the coefficient of the exponential expression.

e2x = -7

4

Now, take the natural log of both sides and simplify to solve for x.

ln (e2x) = ln ( i)

2x(ln(e)) = ln (i)

2x = ln (i)

x    = t ln (i)

 

 

 

Sometimes the methods used to solve an equation introduce an extraneous solution, which is a solution that is correct algebraically, but does not satisfy the conditions of the original equation. One such situation arises when the logarithm is applied to both sides of the equation. Remember that the argument of the logarithm must be positive, so if the number we are evaluating in a logarithmic function is non-positive, there is no output.

Example 16 Solve e2x – ex = 56 for x. Leave all answers as exact values.

Solution: Because we cannot move and combine like terms to have a single term on each side of the equals sign, we will move terms so that we produce an equation that is equal to

zero.

e2x – ex= 56

e2x – ex – 56 = 0

 

While it may not be immediately obvious, e2x – ex – 56 is in quadratic form, (ex)2 –

ex – 56. Recognizing this allows us to factor, and solve for x.

(ex)2 –  ex – 56 = 0

(ex + 7)( ex – 8) = 0

ex + 7 = 0      or      ex – 8 = 0

ex = – 7     or      ex = 8

 

Given ex > 0 for all values of x, then ex -/=- -7, and we only have that ex = 8. Solving for x, we have

ln (ex) = ln(8)

x = ln(8)

If you do not reject the equation ex = -7, as was done in the solution, then you should see the extraneous solution when you get ln (ex) = ln(- 7) as part of the process, which is undefined.

Solving Equations Involving Logarithms

As with exponential equations, we can use the One-to-One Property to solve logarithmic equations.

For example, if

then by the One-to-One Property

x-1=8

X = 9.

While the reader has been encouraged to check their solutions in previous sections, it is im­ perative to do so when solving equations involving logarithms, as these equations often have extraneous solutions. When given an equation with logarithms of the same base on each side, we can use the rules of logarithms to rewrite each side as a single logarithm. Then we use the fact that logarithmic functions are one-to-one to set the arguments equal to one another and

solve for the unknown. For example, consider the equation log(3x – 2) – log(2) = log(x + 4).

 

To solve this equation for x, we can use the rules of logarithms to rewrite the left-hand side as a single logarithm, and then apply the One-to-One Property.

log(3x – 2) – log(2) = log(x + 4)

3

2

log ( x;      ) = log(x + 4)

3x-2 =x+4

2

3x – 2 = 2( X + 4)

3x-    2 = 2x + 8

X = 10

Our original equation contains logarithms, so we must check to make sure this answer is not extraneous.

To check, substitute x = 10 into the original equation: log(3x – 2) – log(2) = log(x + 4).

?

log(3(10) – 2) – log(2) •    log(lO + 4)

?

log(28) – log(2) • log(14) log (228)          7 log(14) log(14) = log(14)

The solution checks and all arguments of logs in the original equation (28, 2, and 14) are positive.

So, the solution to log(3x – 2) – log(2) = log(x + 4) is x = 10.

Note: For x to be a solution to the equation log(3x – 2) – log(2) = log(x + 4), x must be in the intersection of the domains of the logarithms.

The domain for each logarithm in the equation is

log(3x – 2) :    x > l

log(2) :             -oo  < x < oo

{  log(x + 4) :      x > -4

Intersection of the domains: x > i or (i, oo)

Our solution x = 10 > l- Therefore, we know x = 10 is a valid solution to the equation.

Example 17: Solve ln (x2) = ln(2x + 3) for x. Leave all answers as exact values.

Solution: We already have a single logarithm, in the same base, on each side of the equation.

Thus, we can apply the One-to-One Property, and solve for x.

ln (x2) = ln(2x + 3) x2 = 2x + 3

x2 –  2x – 3 = 0

(x – 3)(x + 1) = 0

x – 3 = 0 or      x + 1 = 0

x = 3 or      x = -1

To check, first, substitute x = 3 into the original equation: ln (x2) = ln(2x + 3).

ln (32)  7 ln(2(3) + 3)

ln(9) = ln(9)

Next, substitute x = -1 into the original equation: ln (x2) = ln(2x + 3).

ln ((-1 )2) 7 ln(2(-1)   + 3)

ln(l) = ln(l)

While the solution x = -1 is negative, it checks when substituted into the original equation because the arguments of the logarithmic functions are still positive, as is also the case with x = 3. Thus, x = 3 and x = -1 are both solutions to the equation ln (x2) = ln(2x + 3).

Example 18: Solve log7(1 – 2x) = 1 – log7(3 –  x) for x. Leave all answers as exact values.

Solution: We begin by rewriting the equation with the logarithms of the same base on the same side of the equals sign, which we can then condense to a single logarithm in the same base.

log7(1 – 2x) = 1 – log7(3 – x)

log7(1 – 2x) + log7(3 – x) = 1 log7[(1 – 2x)(3 – x)] = 1

 

Rewriting the logarithmic equation in its equivalent exponential form gives

71 = (1 – 2x) (3 – X)

(1 – 2x)(3 – X) = 7

Now, we can multiply the binomial factors and solve the quadratic equation that results.

2×2 –  7x + 3 = 7

2×2 – 7x – 4 = 0 (2x + 1)(X –  4) = 0

2x + 1 = 0 or      x – 4 = 0

1

x =  —     or      x = 4

2

To check the results, first substitute x =-½into the original equation: log7(1- 2x) = 1 – log7(3 – x).

 

log7 ( 1 – 2 ( –  )  )       1 – log7 ( 3 –   ( – )   )

 

7                                              7

log (1 + 1)) 7 1 – log  (;)

 

?

?

?

log7(2) log7(2)

log7(2)

1 – [log7(7) – log7(2)]

1 – [1 –  log7(2)]

1 – 1 + log7(2)

log7(2) = log7(2)

The solution x = -½ checks and all the logarithms are defined. Next, substitute x = 4 into the original equation: log7( 1 – 2x) = 1 – log7( 3 – x).

We can stop checking at this point, as logarithms cannot have a negative argument, which occurs with log7(-7) and log7(-1). Thus, even though x = 4 is positive, it is an extraneous solution for the original equation. Therefore, the solution to log7 ( 1 – 2x) = 1 – log7(3 – x) is x = -½-

?

Note: When checking a result for validity, if you get, for example, logb(-8)       logb(-8) or

?

logb(0) •     logb(0), the result is NOT a solution, even though the left-hand and right-hand

sides of the equation appear equal. The argument of a logarithm must be positive in order for a solution to be valid.

 

 

 

 

Computing Domains of Algebraic Functions

When computing the domain of an algebraic function, it may be necessary to solve equations involving exponentials and/or logarithms.

 

Example 19: State the domain of the following functions, using interval notation.

(a)    f (X) =  1-:2x

 

(b)   h(x)=   log(3x l)-2

 

Solution: (a) f(x) is the quotient of two functions, so both the numerator and denominator must be defined AND the denominator cannot be zero. The numerator, x, is a linear polynomial and is defined for all real numbers. The denominator, 1- e2x, is an exponential function and is defined for all real numbers. As the denominator, 1 – e2x, cannot be zero. Thus,

1 – e2x-=/=0–

-e2x-=/=- -1

e2x -=/=- 1

ln (e2x) -=/=- ln(l)

2x ln(e) -=/=- ln(l)

2x-=/=- 0

X -=/=- 0

 

So, the domain of f(x) is the intersection of (-00,00) and (-oo,0) U (0,oo). If we drew (-oo, oo) and (-oo, 0) U (0, oo) on a number line (using techniques from Section 5.1), the overlapping segments of these intervals would show that the domain of f(x) is (-oo,0) U (0,oo).

(b) h(x) is the quotient of two functions, so both the numerator and denominator must be defined AND the denominator cannot be zero. The numerator, 5, is a constant polynomial and is defined for all real numbers.  The denominator,

log(3x + 1) – 2, is a logarithmic function and is defined when 3x + 1 is defined

AND when 3x + 1 > 0. As 3x + 1 is a linear polynomial which is defined for all real numbers, we must only determine where 3x + 1 is positive for the logarithm to be defined.

3x + 1 > 0

3x > -1

1

X > —

3

As the denominator, log(3x + 1) – 2 cannot be zero. Thus,

log(3x + 1) – 2-/- 0

log(3x + 1) -/- 2

3x + 1 -/- 102

3x + 1 -/- 100

3x-/- 99

X-/- 33

So, the domain of h(x) is the intersection of these intervals: (-00,00), (-½,oo), and (-oo, 33) U (33, oo). If we drew (-oo, oo), (-½, oo), and (-oo, 33) U (33, oo) on a number line (using techniques from Section 5.1), the overlapping segments of these intervals would show that the domain of h(x) is (-½,33) U (33,oo).

 

 

 

Real-World Applications

Historically, logarithms have played a huge role in the scientific development of our society. They were used to develop analog computing devices called slide rules which enabled scien­ tists and engineers to perform accurate calculations leading to such things as space travel and the moon landing.

 

In a previous section, we learned that exponential functions are used to model exponential growth and decay, including compound interest. When solving an exponential model for a variable in the exponent, we use logarithms.

 

Due to the applied nature of the problems we will examine here, the calculator is often used to express our answers as decimal approximations, after calculating the exact answers.

 

Example 20: Suppose P dollars is invested in an account which offers interest of 7.125% per year.

How long does it take the initial investment to double if

(a)    The interest is compounded monthly?

(b)   The interest is compounded continuously?

Solution:      (a) When interest is compounded m times per year, recall that the amount of money in the account is given by

A(t) = P ( l +mr )mt

 

The initial amount is given to be P.  The annual interest rate is 7.125%, so r = 0.07125. Because we are compounding monthly, m = 12. We want to calculate the amount of time it takes for the account to double, so we are solving

for the value oft when A= 2P. Using this information in the formula and solving for t gives us

2P = P (

1 + 0.07125)

12

12t

2P =P(l.OO59375)12t

2 JJ = ?(1.OO59375)12t

2 = (1.OO59375)12t

ln(2) = ln(l.OO59375)12t ln(2) = 12t ln(l.OO59375)

t =             ln(2)           12 ln(l.OO59375)

t;::::::9.76

Thus, after approximately 9.76 years the account will double.

(b) When interest is compounded continuously, recall the amount of money in the account is given by

A(t) = Pert

 

Using the given values of A and r (the same as part a) and solving fort, we have

2p=   Peo.01125t

2? =   ;Peo.01125t

2=   eo.01125t

ln(2) = ln (eo.01125t) ln(2) = O.O7125t(ln(e))1

ln(2)

t =     0.07125

t;:::; 9.73

 

Thus, after approximately 9.73 years the account will double. (A little less time than when only compounding monthly).

 

Example 21: If the number of people, N(t), in hundreds, at a local community college who have heard the rumor “Adam is afraid of Virginia Woolf,” after t days can be modeled by

N(t) –           84

–      1 + 2799e-t ‘

how long will it be before 4200 people have heard the rumor?

Solution: To determine how long it takes until 4200 people have heard the rumor, we need to solve fort, after substituting the appropriate value for N(t). N(t) is written in hundreds, so

if 4200 people have heard the rumor, then N(t) = 4120°0°=  42. Now, we are ready to

 

solve fort.                                                                 84

42 = 1 + 2799e-t

42 (1 + 2799e-t) = 84

1 + 2799e-t = 2

2799e-t = 1

e-t

 

=–

1

2799

ln (e-t) = ln ( 2;99)

-tln(e) = ln (2;99)

 

-t = ln (2:99)

 

t    =-     ln(  2:99)

t = -[ln(l) – ln(2799)]

t = -[0 – ln(2799)]

t = ln(2799)

t      7.94

Thus, after approximately 8 days, 4200 people would have heard “Adam is afraid of Virginia Woolf.”

Thy It Answers

1.   Check (f o g)(x) = x and (go f)(x) = x.

(f o g)(x) = f(g(x))

(3x – 5) + 5

–

3

3x

–       3                                       =x+5-5

=x                                        =x

 

2.      (a) 106 = 1,000,000

(b)        52 = 25

3.      (a) log3(9) = 2

(b)       log5(125) = 3

(c)        log2 (½) = -1

4. -5

og

•

5 1    (5ytx(x-1)3)

(7x-l)

6.    2 logb(x) + 3 logb(Y) – 4 logb(z) – ½

7.      (a) (0, oo)

(b)        (-oo, oo)

(c)        As x       0 from the right, f(x)        oo As x       oo, f(x)        -oo

(d)   (1, 0)

(e)    No y-intercept

8.    ( ,oo)

9.   t = pog (½)

lo

_ ln(4)

•    X –         ln(8)

11.    X = e

12.    X = 5

13.    Approximately 38.55 years

Exercises 

Basic Skills Practice 

For Exercises 1 – 3, determine if the graph of f(x) is a one-to-one function.

 

1.  

 

2.

 

3.

 

For Exercises 4 – 5, determine if f(x) and g(x) are inverse functions.

4.     f(x) = 6x – 2 and g(x) = ¼x + ½                  5.  f (x) = 42 –  x and g(x) = 42 + x

 

For Exercises 6 – 9, write the logarithmic equation in exponential form.

6.   log4(16) = 2

7.   log3(81) = 4

8.   log C6o) = -2

9.   ln( ve) = ½

For Exercises 10 – 13, write the exponential equation in logarithmic form.

 

10.    e0 = 1

11.   105 = 100000

12.   5-3 = 1 5

13.    2-6  = l4

For Exercises 14 – 17, express each as a single logarithm. Assume when necessary that all variables represent positive real numbers.

 

14.    2log8(x) – 4log8(y)

15.    – log5(2) + 11 log5(y)

16.   3ln(2) + 7ln(x)

17.    y12log(y) – 3log(x)

For Exercises 18 – 21, use the properties of logarithms to fully expand and simplify each expression. Assume when necessary that all variables represent positive real numbers and y > z.

18.    log(lO0xyz)                                                     20. log7 (xy2 😉

 

21.      log13 ( 13;tY)

 

For Exercises 22 – 23, state the following properties of the logarithmic function, if they exist.

 

(a)   Domain

(b)   Range

(c) End behavior

22.         f(x) = log8(x)

(d)       x – intercept(s)

e. y – intercept(s)

 

 

23.      g(x) = log1(x)

8

For Exercises 24 – 27, state the domain of each logarithmic function, using interval notation.

 

24.      f (x) = log(x – 5)

25.       h(x) = ln(2x – 7)

26.      g(x) = log3(x + 9)

27.        j(x) = log17 (3x + 8)

For Exercises 28 – 31, solve each of the following for x. Leave your answers in exact form.

28.      5x = 7

29.       102x-9  = 3                                                       31. o.4s-3x = 11

 

For Exercises 32 – 37, solve each of the following for x. Leave your answers in exact form.

 

32.      log2(x) = 4

33.      5ln(x)=l

34.      log(4x – 5) = 3

35.      ln(2 – 3x) – 1 = 0

36.      log2 (x2 –  8x – 25) = 3

37.      log4(x + 6) – log4(x – 2) = 1

For Exercises 38 – 39, the model for continuous (exponential) growth/ decay is given by y = cekt, where c is the initial amount, k is the relative growth rate (as a decimal), tis time (in years), and y is the future amount after t years.

38.       At the beginning of the year 2000, the world population was about six billion people. Assuming the population increases continuously at a rate of 1.3% per year, how many years later will the population reach eight billion people? Round your answer to the nearest tenth of a year.

39.      A new piece of equipment worth $75,000 depreciates continuously at a relative rate of 8.2% per year. How long will it take for the equipment to reach its scrap value of

$23, 0007 Round your answer to three decimal places.

 

 

 

Intermediate Skills Practice

 

For Exercises 40 – 43, determine if f(x) is a one-to-one function, using the Horizontal Line Test.

 

40.         f ( X) = 3x + 5

41.         f(x) = 2lx –  71

42.         f(x) = 5×3 –  x2 + 9

43.         f(x) = e-x

For Exercises 44 – 45, determine if f(x) and g(x) are inverse functions.

44.

4                                                                                                                     1 3

J(x) =    x and g(x) = 4x;3                            45. f(x) =       x and g(x) = For Exercises 46 – 51, evaluate each expression, without using a calculator.

46.       ln (e3)

47.       log5( v15)

49.      log19 (3 1)

50.      1010g(v’3)

 

 

 

For Exercises 52 – 54, express each as a single logarithm. Assume when necessary that all variables represent positive real numbers.

52. log(x) – 3log(y) – ½ log(z)

53. ½ ln(x) – ln(x + y) + 4ln(2z)

For Exercises 55 – 58, use the properties of logarithms to fully expand and simplify each expression. Assume when necessary that all variables represent positive real numbers greater than 2.

 

 

55•

log3 ( yx8+..fz5)

57. log5 J625xy3

 

56.      ln(  )                                                               58. log2(x(x + 4)(2x – 3))

For Exercises 59 – 60, state the following properties of the logarithmic function, if they exist.

 

(a)   Domain

(b)   Range

(c)  End behavior

59.         f(x) = log(x)

(d)   x – intercept(s)

e. y – intercept(s)

 

 

60.      g(x) = ln(x)

For Exercises 61 – 66, state the domain of the algebraic function, using interval notation.

 

61.         f(x) = ln(-6x + 11)

62.       j(x) = log2(10+ 33x) + 45

63.      g(x) = log (\]’x + 4)

3

65. h(x)=  log(9-x)

x2 -49

 

 

=     log(x + 1)

64. k(x) = ln (ex) – 35

66.      m (X )

 

 

1v1xr::–::8-o

For Exercises 67 – 72, solve each of the following for x. Leave your answers in exact form.

 

67.      4e-3x = 5

68.      53+5x –  12 = 0

69.       2 · 3-x = 16

70.      e4x = 5e7x

71.      11vx = 21

72.      ex2 = 14

For Exercises 73 – 76, solve each of the following for x. Leave your answers in exact form.

 

73.      (2x – 5) (2x – 13) = 0

74.      (78x + 14) (92x – 36) = 0

75.      (3x – 9) (3x – 4) = 0

76.      (ex+1)(3ex-2)=0

 

For Exercises 77 – 82, solve each of the following for x. Leave your answers in exact form.

 

77.      log(x) + log(x – 5) = log(24)

78.      log5(2x – 6) – 3 = 0

79.      log8(4- x) = log8(12) – log8(-x)

80.      log(14x + 3) = 2 + log(x – 1)

81.      ln(x – 1) = ln(l) – ln(3x)

82.      log7(2x – 3) – log7(x + 1) = log7(10)

83.       If $3600 is invested in an account that earns interest at an annual rate of 4.7%, com­ pounded continuously, how long will it take for the account to reach $12, 000? Round your answer to the nearest year.

84.      At what annual interest rate must you invest $1000 in order to have $100,000 in 45 years, assuming the account is compounded continuously? Round your percentage to four decimal places.

 

 

 

Mastery Practice

 

 

83.

 

 

84.

 

Given J(x) = {

 

 

Given J(x) = {

x2          if X < 0

2                                if X = 0 , determine if f (x) is a one-to-one function.

-2x-3      if X > 0

x2          if X < 0

-2           if X = 0 , determine if f (x) is a one-to-one function.

-2x-3      if X > 0

85.       Determine if f(x) = x2- lOx, for x 2:: 5, and g(x) = 5 + Jx + 25 are inverse functions.

86.       Express the given expression as a single logarithm, assuming all variables represent positive real numbers.

 

87.       Expand and simplify the given expression. Assume when necessary that all variables represent positive real numbers greater than 3.

(x – 3)y7/s)

logb(          bz

 

88.      State the following properties of the logarithmic function f (x) = loga (x – 4) + 2 for real number a > 1.

 

(a)   Domain

(b)   Range

(c) End behavior

(d) x – intercept(s)

e. y – intercept(s)

 

For Exercises 91 – 94, state the domain of each algebraic function, using interval notation.

evx+2

91.

3

J(x) = log1 ( 4 – x)

h()             4-3x

•

92

x  =     log5(x + 7) – 1

ln(-x)

93.

X

g(x) = 2- e3 +9

94    .( )        log(x + 2) – log(8 – x)

•      J X  =

-7 +  lQ2x-1