6.1 Introduction to Exponential and Logarithmic Functions

Of all of the functions we study in this text, exponential and logarithmic functions are possibly the ones which impact everyday life the most.¹ This section introduces us to these functions while the rest of the chapter will more thoroughly explore their properties. Up to this point, we have dealt with functions which involve terms like $x^{2}$ or $x to the power 2 / 3$ , in other words, terms of the form $x^{p}$ where the base of the term, $x$ , varies but the exponent of each term, $p$ , remains constant. In this chapter, we study functions of the form $f (x) = b^{x}$ where the base $b$ is a constant and the exponent $x$ is the variable. We start our exploration of these functions with $f (x) = 2^{x}$ . (Apparently this is a tradition. Every College Algebra book we have ever read starts with $f (x) = 2^{x}$ .) We make a table of values, plot the points and connect the dots in a pleasing fashion.

$\begin{matrix} x & f (x) & (x, f (x)) \\ - 3 & 2^{- 3} = \frac{1}{8} & (- 3, \frac{1}{8}) \\ - 2 & 2^{- 2} = \frac{1}{4} & (- 2, \frac{1}{4}) \\ - 1 & 2^{- 1} = \frac{1}{2} & (- 1, \frac{1}{2}) \\ 0 & 2^{0} = 1 & (0, 1) \\ 1 & 2^{1} = 2 & (1, 2) \\ 2 & 2^{2} = 4 & (2, 4) \\ 3 & 2^{3} = 8 & (3, 8) \end{matrix}$

Coordinate-plane figure. — **Figure 6.1** $y = f (x) = 2^{x}$

A few remarks about the graph of $f (x) = 2^{x}$ which we have constructed are in order. As $x → minus ∞$ and attains values like $x = - 100$ or $x = - 1000$ , the function $f (x) = 2^{x}$ takes on values like $f (- 100) = 2^{- 100} = \frac{1}{2^{100}}$ or $f (- 1000) = 2^{- 1000} = \frac{1}{2^{1000}}$ . In other words, as $x → minus ∞$ ,

$2^{x} \approx \frac{1}{very big (+)} \approx very small (+)$

So as $x → minus ∞$ , $2^{x} \to 0^{+}$ . This is represented graphically using the $x$ -axis (the line $y equals 0$ ) as a horizontal asymptote. On the flip side, as $x → ∞$ , we find $f (100) = 2^{100}$ , $f (1000) = 2^{1000}$ , and so on, thus $2 to the power x → ∞$ . As a result, our graph suggests the range of $f$ is $open parenthesis 0, ∞ close parenthesis$ . The graph of $f$ passes the Horizontal Line Test which means $f$ is one-to-one and hence invertible. We also note that when we `connected the dots in a pleasing fashion', we have made the implicit assumption that $f (x) = 2^{x}$ is continuous² and has a domain of all real numbers. In particular, we have suggested that things like $2^{\sqrt{3}}$ exist as real numbers. We should take a moment to discuss what something like $2^{\sqrt{3}}$ might mean, and refer the interested reader to a solid course in Calculus for a more rigorous explanation. The number $\sqrt{3} = 1.73205 \dots$ is an irrational number³ and as such, its decimal representation neither repeats nor terminates. We can, however, approximate $the square root of 3$ by terminating decimals, and it stands to reason⁴ we can use these to approximate $2^{\sqrt{3}}$ . For example, if we approximate $the square root of 3$ by $1.73$ , we can approximate $2^{\sqrt{3}} \approx 2^{1.73} = 2^{\frac{173}{100}} = \sqrt[100]{2^{173}}$ . It is not, by any means, a pleasant number, but it is at least a number that we understand in terms of powers and roots. It also stands to reason that better and better approximations of $the square root of 3$ yield better and better approximations of $2^{\sqrt{3}}$ , so the value of $2^{\sqrt{3}}$ should be the result of this sequence of approximations.⁵

Suppose we wish to study the family of functions $f (x) = b^{x}$ . Which bases $b$ make sense to study? We find that we run into difficulty if $b less than 0$ . For example, if $b = - 2$ , then the function $f (x) = {(- 2)}^{x}$ has trouble, for instance, at $x = \frac{1}{2}$ since ${(- 2)}^{1 / 2} = \sqrt{- 2}$ is not a real number. In general, if $x$ is any rational number with an even denominator, then ${(- 2)}^{x}$ is not defined, so we must restrict our attention to bases $b greater than or equal to 0$ . What about $b equals 0$ ? The function $f (x) = 0^{x}$ is undefined for $x less than or equal to 0$ because we cannot divide by $0$ and $0 to the power 0$ is an indeterminant form. For $x greater than 0$ , $0^{x} = 0$ so the function $f (x) = 0^{x}$ is the same as the function $f ⁢ open parenthesis x close parenthesis equals 0$ , $x greater than 0$ . We know everything we can possibly know about this function, so we exclude it from our investigations. The only other base we exclude is $b equals 1$ , since the function $f (x) = 1^{x} = 1$ is, once again, a function we have already studied. We are now ready for our definition of exponential functions.

We leave it to the reader to verify⁶ that if $b greater than 1$ , then the exponential function $f (x) = b^{x}$ will share the same basic shape and characteristics as $f (x) = 2^{x}$ . What if $0 less than b less than 1$ ? Consider $g (x) = {(\frac{1}{2})}^{x}$ . We could certainly build a table of values and connect the points, or we could take a step back and note that $g (x) = {(\frac{1}{2})}^{x} = {(2^{- 1})}^{x} = 2^{- x} = f (- x)$ , where $f (x) = 2^{x}$ . Thinking back to Section, the graph of $f (- x)$ is obtained from the graph of $f (x)$ by reflecting it across the $y$ -axis. We get

$\overset{\overset{reflect across y -axis}{\overset{}{\to}}}{multiply each x -coordinate by - 1}$

We see that the domain and range of $g$ match that of $f$ , namely $open parenthesis minus ∞, ∞ close parenthesis$ and $open parenthesis 0, ∞ close parenthesis$ , respectively. Like $f$ , $g$ is also one-to-one. Whereas $f$ is always increasing, $g$ is always decreasing. As a result, as $x → minus ∞$ , $g ⁢ open parenthesis x close parenthesis → ∞$ , and on the flip side, as $x → ∞$ , $g (x) \to 0^{+}$ . It shouldn't be too surprising that for all choices of the base $0 less than b less than 1$ , the graph of $y = b^{x}$ behaves similarly to the graph of $g$ . We summarize the basic properties of exponential functions in the following theorem.⁷

Theorem

Properties of Exponential Functions: Suppose $f (x) = b^{x}$ .

The domain of $f$ is $open parenthesis minus ∞, ∞ close parenthesis$ and the range of $f$ is $open parenthesis 0, ∞ close parenthesis$ .
$open parenthesis 0, 1 close parenthesis$ is on the graph of $f$ and $y equals 0$ is a horizontal asymptote to the graph of $f$ .
$f$ is one-to-one, continuous and smooth⁸

If $b greater than 1$ :
- $f$ is always increasing
- As $x → minus ∞$ , $f (x) \to 0^{+}$
- As $x → ∞$ , $f ⁢ open parenthesis x close parenthesis → ∞$
- The graph of $f$ resembles:
  Figure 6.4 $y = b^{x}$ , $b greater than 1$

If $0 less than b less than 1$ :
- $f$ is always decreasing
- As $x → minus ∞$ , $f ⁢ open parenthesis x close parenthesis → ∞$
- As $x → ∞$ , $f (x) \to 0^{+}$
- The graph of $f$ resembles:
  Figure 6.5 $y = b^{x}$ , $0 less than b less than 1$

Of all of the bases for exponential functions, two occur the most often in scientific circles. The first, base $10$ , is often called the common base. The second base is an irrational number, $e approximately equals 2.718$ , called the natural base. We will more formally discuss the origins of this number in Section. For now, it is enough to know that since $e greater than 1$ , $f (x) = e^{x}$ is an increasing exponential function. The following examples give us an idea how these functions are used in the wild.

Example 1

The value of a car can be modeled by $V (x) = 25 {(\frac{4}{5})}^{x}$ , where $x greater than or equal to 0$ is age of the car in years and $V (x)$ is the value in thousands of dollars.

Find and interpret $V ⁢ open parenthesis 0 close parenthesis$ .
Sketch the graph of $y = V (x)$ using transformations.
Find and interpret the horizontal asymptote of the graph you found in 2.

To find $V ⁢ open parenthesis 0 close parenthesis$ , we replace $x$ with $0$ to obtain $V (0) = 25 {(\frac{4}{5})}^{0} = 25$ . Since $x$ represents the age of the car in years, $x equals 0$ corresponds to the car being brand new. Since $V (x)$ is measured in thousands of dollars, $V ⁢ open parenthesis 0 close parenthesis equals 25$ corresponds to a value of $$ 25, 000$ . Putting it all together, we interpret $V ⁢ open parenthesis 0 close parenthesis equals 25$ to mean the purchase price of the car was $$ 25, 000$ .
To graph $y = 25 {(\frac{4}{5})}^{x}$ , we start with the basic exponential function $f (x) = {(\frac{4}{5})}^{x}$ . Since the base $b = \frac{4}{5}$ is between $0$ and $1$ , the graph of $y = f (x)$ is decreasing. We plot the $y$ -intercept $open parenthesis 0, 1 close parenthesis$ and two other points, $(- 1, \frac{5}{4})$ and $(1, \frac{4}{5})$ , and label the horizontal asymptote $y equals 0$ . To obtain $V (x) = 25 {(\frac{4}{5})}^{x}$ , $x greater than or equal to 0$ , we multiply the output from $f$ by $25$ , in other words, $V ⁢ open parenthesis x close parenthesis equals 25 ⁢ f ⁢ open parenthesis x close parenthesis$ . In accordance with Theorem, this results in a vertical stretch by a factor of $25$ . We multiply all of the $y$ values in the graph by $25$ (including the $y$ value of the horizontal asymptote) and obtain the points $(- 1, \frac{125}{4})$ , $open parenthesis 0, 25 close parenthesis$ and $open parenthesis 1, 20 close parenthesis$ . The horizontal asymptote remains $y equals 0$ . Finally, we restrict the domain to $open bracket 0, ∞ close parenthesis$ to fit with the applied domain given to us. We have the result below.
Figure 6.6 $y = f (x) = {(\frac{4}{5})}^{x}$
$\overset{\overset{vertical scale by a factor of 25}{\overset{}{\to}}}{multiply each y -coordinate by 25}$
Figure 6.7 $y = V (x) = 25 f (x)$ , $x greater than or equal to 0$
We see from the graph of $V$ that its horizontal asymptote is $y equals 0$ . (We leave it to reader to verify this analytically by thinking about what happens as we take larger and larger powers of $\frac{4}{5}$ .) This means as the car gets older, its value diminishes to $0$ .

The function in the previous example is often called a `decay curve'. Increasing exponential functions are used to model `growth curves' and we shall see several different examples of those in Section. For now, we present another common decay curve which will serve as the basis for further study of exponential functions. Although it may look more complicated than the previous example, it is actually just a basic exponential function which has been modified by a few transformations from Section.

Example 2

According to Newton's Law of Cooling ⁹ the temperature of coffee $T$ (in degrees Fahrenheit) $t$ minutes after it is served can be modeled by $T (t) = 70 + 90 e^{- 0.1 t}$ .

Find and interpret $T ⁢ open parenthesis 0 close parenthesis$ .
Sketch the graph of $y = T (t)$ using transformations.
Find and interpret the horizontal asymptote of the graph.

To find $T ⁢ open parenthesis 0 close parenthesis$ , we replace every occurrence of the independent variable $t$ with $0$ to obtain $T (0) = 70 + 90 e^{- 0.1 (0)} = 160$ . This means that the coffee was served at $160 degrees ⁢ F$ .
To graph $y = T (t)$ using transformations, we start with the basic function, $f (t) = e^{t}$ . As we have already remarked, $e approximately equals 2.718 greater than 1$ so the graph of $f$ is an increasing exponential with $y$ -intercept $open parenthesis 0, 1 close parenthesis$ and horizontal asymptote $y equals 0$ . The points $(- 1, e^{- 1}) \approx (- 1, 0.37)$ and $open parenthesis 1, e close parenthesis approximately equals open parenthesis 1, 2.72 close parenthesis$ are also on the graph. Since the formula $T (t)$ looks rather complicated, we rewrite $T (t)$ in the form presented in Theorem and use that result to track the changes to our three points and the horizontal asymptote. We have
$T (t) = 70 + 90 e^{- 0.1 t} = 90 e^{- 0.1 t} + 70 = 90 f (- 0.1 t) + 70$
Multiplication of the input to $f$ , $t$ , by $minus 0.1$ results in a horizontal expansion by a factor of $10$ as well as a reflection about the $y$ -axis. We divide each of the $x$ values of our points by $minus 0.1$ (which amounts to multiplying them by $minus 10$ ) to obtain $(10, e^{- 1})$ , $open parenthesis 0, 1 close parenthesis$ , and $open parenthesis minus 10, e close parenthesis$ . Since none of these changes affected the $y$ values, the horizontal asymptote remains $y equals 0$ . Next, we see that the output from $f$ is being multiplied by $90$ . This results in a vertical stretch by a factor of $90$ . We multiply the $y$ -coordinates by $90$ to obtain $(10, 90 e^{- 1})$ , $open parenthesis 0, 90 close parenthesis$ , and $open parenthesis minus 10, 90 ⁢ e close parenthesis$ . We also multiply the $y$ value of the horizontal asymptote $y equals 0$ by $90$ , and it remains $y equals 0$ . Finally, we add $70$ to all of the $y$ -coordinates, which shifts the graph upwards to obtain $(10, 90 e^{- 1} + 70) \approx (10, 103.11)$ , $open parenthesis 0, 160 close parenthesis$ , and $(- 10, 90 e + 70) \approx (- 10, 314.64)$ . Adding $70$ to the horizontal asymptote shifts it upwards as well to $y equals 70$ . We connect these three points using the same shape in the same direction as in the graph of $f$ and, last but not least, we restrict the domain to match the applied domain $open bracket 0, ∞ close parenthesis$ . The result is below.
Figure 6.8 $y = f (t) = e^{t}$
$\overset{}{\to}$
Figure 6.9 $y = T (t)$
From the graph, we see that the horizontal asymptote is $y equals 70$ . It is worth a moment or two of our time to see how this happens analytically and to review some of the `number sense' developed in Chapter. As $t → ∞$ , We get $T (t) = 70 + 90 e^{- 0.1 t} \approx 70 + 90 e^{very big (-)}$ . Since $e greater than 1$ ,
$e^{very big (-)} = \frac{1}{e^{very big (+)}} \approx \frac{1}{very big (+)} \approx very small (+)$
The larger $t$ becomes, the smaller $e^{- 0.1 t}$ becomes, so the term $90 e^{- 0.1 t} \approx very small (+)$ . Hence, $T (t) \approx 70 + very small (+)$ which means the graph is approaching the horizontal line $y equals 70$ from above. This means that as time goes by, the temperature of the coffee is cooling to $70 degrees ⁢ F$ , presumably room temperature.

As we have already remarked, the graphs of $f (x) = b^{x}$ all pass the Horizontal Line Test. Thus the exponential functions are invertible. We now turn our attention to these inverses, the logarithmic functions, which are called `logs' for short.

Definition

The inverse of the exponential function $f (x) = b^{x}$ is called the base $b$ logarithm function, and is denoted $f^{- 1} (x) = \log_{b} (x)$ We read ` $\log_{b} (x)$ ' as `log base $b$ of $x$ .'

We have special notations for the common base, $b equals 10$ , and the natural base, $b = e$ .

Definition

The common logarithm of a real number $x$ is $log sub 10 ⁡ open parenthesis x close parenthesis$ and is usually written $\log (x)$ . The natural logarithm of a real number $x$ is $\log_{e} (x)$ and is usually written $\ln (x)$ .

Since logs are defined as the inverses of exponential functions, we can use Theorems and to tell us about logarithmic functions. For example, we know that the domain of a log function is the range of an exponential function, namely $open parenthesis 0, ∞ close parenthesis$ , and that the range of a log function is the domain of an exponential function, namely $open parenthesis minus ∞, ∞ close parenthesis$ . Since we know the basic shapes of $y = f (x) = b^{x}$ for the different cases of $b$ , we can obtain the graph of $y = f^{- 1} (x) = \log_{b} (x)$ by reflecting the graph of $f$ across the line $y = x$ as shown below. The $y$ -intercept $open parenthesis 0, 1 close parenthesis$ on the graph of $f$ corresponds to an $x$ -intercept of $open parenthesis 1, 0 close parenthesis$ on the graph of $f^{- 1}$ . The horizontal asymptotes $y equals 0$ on the graphs of the exponential functions become vertical asymptotes $x equals 0$ on the log graphs.

$\XYZfig 10$

On a procedural level, logs undo the exponentials. Consider the function $f (x) = 2^{x}$ . When we evaluate $f (3) = 2^{3} = 8$ , the input $3$ becomes the exponent on the base $2$ to produce the real number $8$ . The function $f^{- 1} (x) = \log_{2} (x)$ then takes the number $8$ as its input and returns the exponent $3$ as its output. In symbols, $log sub 2 ⁡ open parenthesis 8 close parenthesis equals 3$ . More generally, $log sub 2 ⁡ open parenthesis x close parenthesis$ is the exponent you put on $2$ to get $x$ . Thus, $log sub 2 ⁡ open parenthesis 16 close parenthesis equals 4$ , because $2^{4} = 16$ . The following theorem summarizes the basic properties of logarithmic functions, all of which come from the fact that they are inverses of exponential functions.

Theorem

Properties of Logarithmic Functions: Suppose $f (x) = \log_{b} (x)$ .

The domain of $f$ is $open parenthesis 0, ∞ close parenthesis$ and the range of $f$ is $open parenthesis minus ∞, ∞ close parenthesis$ .
$open parenthesis 1, 0 close parenthesis$ is on the graph of $f$ and $x equals 0$ is a vertical asymptote of the graph of $f$ .
$f$ is one-to-one, continuous and smooth
$b^{a} = c$ if and only if $\log_{b} (c) = a$ . That is, $\log_{b} (c)$ is the exponent you put on $b$ to obtain $c$ .
$\log_{b} (b^{x}) = x$ for all $x$ and $b^{\log_{b} (x)} = x$ for all $x greater than 0$

If $b greater than 1$ :
- $f$ is always increasing
- As $x \to 0^{+}$ , $f ⁢ open parenthesis x close parenthesis → minus ∞$
- As $x → ∞$ , $f ⁢ open parenthesis x close parenthesis → ∞$
- The graph of $f$ resembles:
  Figure 6.11

If $0 less than b less than 1$ :
- $f$ is always decreasing
- As $x \to 0^{+}$ , $f ⁢ open parenthesis x close parenthesis → ∞$
- As $x → ∞$ , $f ⁢ open parenthesis x close parenthesis → minus ∞$
- The graph of $f$ resembles:
  Figure 6.12

As we have mentioned, Theorem is a consequence of Theorems and. However, it is worth the reader's time to understand Theorem from an exponential perspective. For instance, we know that the domain of $g ⁢ open parenthesis x close parenthesis equals log sub 2 ⁡ open parenthesis x close parenthesis$ is $open parenthesis 0, ∞ close parenthesis$ . Why? Because the range of $f (x) = 2^{x}$ is $open parenthesis 0, ∞ close parenthesis$ . In a way, this says everything, but at the same time, it doesn't. For example, if we try to find $log sub 2 ⁡ open parenthesis minus 1 close parenthesis$ , we are trying to find the exponent we put on $2$ to give us $minus 1$ . In other words, we are looking for $x$ that satisfies $2^{x} = - 1$ . There is no such real number, since all powers of $2$ are positive. While what we have said is exactly the same thing as saying `the domain of $g ⁢ open parenthesis x close parenthesis equals log sub 2 ⁡ open parenthesis x close parenthesis$ is $open parenthesis 0, ∞ close parenthesis$ because the range of $f (x) = 2^{x}$ is $open parenthesis 0, ∞ close parenthesis$ ', we feel it is in a student's best interest to understand the statements in Theorem at this level instead of just merely memorizing the facts.

Example 3

Simplify the following.

$log sub 3 ⁡ open parenthesis 81 close parenthesis$
$\log_{2} (\frac{1}{8})$
$log sub the square root of 5 ⁡ open parenthesis 25 close parenthesis$
$\ln (\sqrt[3]{e^{2}})$
$log ⁡ open parenthesis 0.001 close parenthesis$
$2 to the power log sub 2 ⁡ open parenthesis 8 close parenthesis$
$117^{- \log_{117} (6)}$

The number $log sub 3 ⁡ open parenthesis 81 close parenthesis$ is the exponent we put on $3$ to get $81$ . As such, we want to write $81$ as a power of $3$ . We find $81 = 3^{4}$ , so that $log sub 3 ⁡ open parenthesis 81 close parenthesis equals 4$ .
To find $\log_{2} (\frac{1}{8})$ , we need rewrite $\frac{1}{8}$ as a power of $2$ . We find $\frac{1}{8} = \frac{1}{2^{3}} = 2^{- 3}$ , so $\log_{2} (\frac{1}{8}) = - 3$ .
To determine $log sub the square root of 5 ⁡ open parenthesis 25 close parenthesis$ , we need to express $25$ as a power of $the square root of 5$ . We know $25 equals 5 squared$ , and $5 = {(\sqrt{5})}^{2}$ , so we have $25 = {({(\sqrt{5})}^{2})}^{2} = {(\sqrt{5})}^{4}$ . We get $\log_{\sqrt{5}} (25) = 4$ .
First, recall that the notation $\ln (\sqrt[3]{e^{2}})$ means $\log_{e} (\sqrt[3]{e^{2}})$ , so we are looking for the exponent to put on $e$ to obtain $\sqrt[3]{e^{2}}$ . Rewriting $\sqrt[3]{e^{2}} = e^{2 / 3}$ , we find $\ln (\sqrt[3]{e^{2}}) = \ln (e^{2 / 3}) = \frac{2}{3}$ .
Rewriting $log ⁡ open parenthesis 0.001 close parenthesis$ as $log sub 10 ⁡ open parenthesis 0.001 close parenthesis$ , we see that we need to write $0.001$ as a power of $10$ . We have $0.001 = \frac{1}{1000} = \frac{1}{10^{3}} = 10^{- 3}$ . Hence, $\log (0.001) = \log (10^{- 3}) = - 3$ .
We can use Theorem directly to simplify $2^{\log_{2} (8)} = 8$ . We can also understand this problem by first finding $log sub 2 ⁡ open parenthesis 8 close parenthesis$ . By definition, $log sub 2 ⁡ open parenthesis 8 close parenthesis$ is the exponent we put on $2$ to get $8$ . Since $8 equals 2 cubed$ , we have $log sub 2 ⁡ open parenthesis 8 close parenthesis equals 3$ . We now substitute to find $2^{\log_{2} (8)} = 2^{3} = 8$ .
From Theorem, we know $117^{\log_{117} (6)} = 6$ , but we cannot directly apply this formula to the expression $117^{- \log_{117} (6)}$ . (Can you see why?) At this point, we use a property of exponents followed by Theorem to get¹⁰
$117^{- \log_{117} (6)} = \frac{1}{117^{\log_{117} (6)}} = \frac{1}{6}$

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 logs, we now have another restriction. Since the domain of $f (x) = \log_{b} (x)$ is $open parenthesis 0, ∞ close parenthesis$ , the argument¹¹ of the log must be strictly positive.

Example 4

Find the domain of the following functions. Check your answers graphically using the calculator.

$f (x) = 2 \log (3 - x) - 1$
$g (x) = \ln (\frac{x}{x - 1})$

We set $3 minus x greater than 0$ to obtain $x less than 3$ , or $open parenthesis minus ∞, 3 close parenthesis$ . The graph from the calculator below verifies this. Note that we could have graphed $f$ using transformations. Taking a cue from Theorem, we rewrite $f (x) = 2 \log_{10} (- x + 3) - 1$ and find the main function involved is $y = h (x) = \log_{10} (x)$ . We select three points to track, $(\frac{1}{10}, - 1)$ , $open parenthesis 1, 0 close parenthesis$ and $open parenthesis 10, 1 close parenthesis$ , along with the vertical asymptote $x equals 0$ . Since $f (x) = 2 h (- x + 3) - 1$ , Theorem tells us that to obtain the destinations of these points, we first subtract $3$ from the $x$ -coordinates (shifting the graph left $3$ units), then divide (multiply) by the $x$ -coordinates by $minus 1$ (causing a reflection across the $y$ -axis). These transformations apply to the vertical asymptote $x equals 0$ as well. Subtracting $3$ gives us $x = - 3$ as our asymptote, then multplying by $minus 1$ gives us the vertical asymptote $x equals 3$ . Next, we multiply the $y$ -coordinates by $2$ which results in a vertical stretch by a factor of $2$ , then we finish by subtracting $1$ from the $y$ -coordinates which shifts the graph down $1$ unit. We leave it to the reader to perform the indicated arithmetic on the points themselves and to verify the graph produced by the calculator below.
To find the domain of $g$ , we need to solve the inequality $\frac{x}{x - 1} > 0$ . As usual, we proceed using a sign diagram. If we define $r (x) = \frac{x}{x - 1}$ , we find $r$ is undefined at $x equals 1$ and $r ⁢ open parenthesis x close parenthesis equals 0$ when $x equals 0$ . Choosing some test values, we generate the sign diagram below.
Figure 6.13
We find $\frac{x}{x - 1} > 0$ on $open parenthesis minus ∞, 0 close parenthesis union open parenthesis 1, ∞ close parenthesis$ to get the domain of $g$ . The graph of $y = g (x)$ confirms this. We can tell from the graph of $g$ that it is not the result of Section transformations being applied to the graph $y = \ln (x)$ , so barring a more detailed analysis using Calculus, the calculator graph is the best we can do. One thing worthy of note, however, is the end behavior of $g$ . The graph suggests that as $x \to \pm \infty$ , $g ⁢ open parenthesis x close parenthesis → 0$ . We can verify this analytically. Using results from Chapter and continuity, we know that as $x \to \pm \infty$ , $\frac{x}{x - 1} \approx 1$ . Hence, it makes sense that $g (x) = \ln (\frac{x}{x - 1}) \approx \ln (1) = 0$ .

Figure 6.14

Figure 6.15

$y = f (x) = 2 \log (3 - x) - 1$

$y = g (x) = \ln (\frac{x}{x - 1})$

While logarithms have some interesting applications of their own which you'll explore in the exercises, their primary use to us will be to undo exponential functions. (This is, after all, how they were defined.) Our last example solidifies this and reviews all of the material in the section.

Example 5

Let $f (x) = 2^{x - 1} - 3$ .

Graph $f$ using transformations and state the domain and range of $f$ .
Explain why $f$ is invertible and find a formula for $f^{- 1} (x)$ .
Graph $f^{- 1}$ using transformations and state the domain and range of $f^{- 1}$ .
Verify $(f^{- 1} \circ f) (x) = x$ for all $x$ in the domain of $f$ and $(f \circ f^{- 1}) (x) = x$ for all $x$ in the domain of $f^{- 1}$ .
Graph $f$ and $f^{- 1}$ on the same set of axes and check the symmetry about the line $y = x$ .

If we identify $g (x) = 2^{x}$ , we see $f (x) = g (x - 1) - 3$ . We pick the points $(- 1, \frac{1}{2})$ , $open parenthesis 0, 1 close parenthesis$ and $open parenthesis 1, 2 close parenthesis$ on the graph of $g$ along with the horizontal asymptote $y equals 0$ to track through the transformations. By Theorem we first add $1$ to the $x$ -coordinates of the points on the graph of $g$ (shifting $g$ to the right $1$ unit) to get $(0, \frac{1}{2})$ , $open parenthesis 1, 1 close parenthesis$ and $open parenthesis 2, 2 close parenthesis$ . The horizontal asymptote remains $y equals 0$ . Next, we subtract $3$ from the $y$ -coordinates, shifting the graph down $3$ units. We get the points $(0, - \frac{5}{2})$ , $open parenthesis 1, minus 2 close parenthesis$ and $open parenthesis 2, minus 1 close parenthesis$ with the horizontal asymptote now at $y = - 3$ . Connecting the dots in the order and manner as they were on the graph of $g$ , we get the graph below. We see that the domain of $f$ is the same as $g$ , namely $open parenthesis minus ∞, ∞ close parenthesis$ , but that the range of $f$ is $open parenthesis minus 3, ∞ close parenthesis$ .
Figure 6.16 $y = h (x) = 2^{x}$
$\overset{}{\to}$
Figure 6.17 $y = f (x) = 2^{x - 1} - 3$
The graph of $f$ passes the Horizontal Line Test so $f$ is one-to-one, hence invertible. To find a formula for $f^{- 1} (x)$ , we normally set $y = f (x)$ , interchange the $x$ and $y$ , then proceed to solve for $y$ . Doing so in this situation leads us to the equation $x = 2^{y - 1} - 3$ . We have yet to discuss how to solve this kind of equation, so we will attempt to find the formula for $f^{- 1}$ from a procedural perspective. If we break $f (x) = 2^{x - 1} - 3$ into a series of steps, we find $f$ takes an input $x$ and applies the steps
1. subtract $1$
2. put as an exponent on $2$
3. subtract $3$
Clearly, to undo subtracting $1$ , we will add $1$ , and similarly we undo subtracting $3$ by adding $3$ . How do we undo the second step? The answer is we use the logarithm. By definition, $log sub 2 ⁡ open parenthesis x close parenthesis$ undoes exponentiation by $2$ . Hence, $f^{- 1}$ should
1. add $3$
2. take the logarithm base $2$
3. add $1$
In symbols, $f^{- 1} (x) = \log_{2} (x + 3) + 1$ .
To graph $f^{- 1} (x) = \log_{2} (x + 3) + 1$ using transformations, we start with $j ⁢ open parenthesis x close parenthesis equals log sub 2 ⁡ open parenthesis x close parenthesis$ . We track the points $(\frac{1}{2}, - 1)$ , $open parenthesis 1, 0 close parenthesis$ and $open parenthesis 2, 1 close parenthesis$ on the graph of $j$ along with the vertical asymptote $x equals 0$ through the transformations using Theorem. Since $f^{- 1} (x) = j (x + 3) + 1$ , we first subtract $3$ from each of the $x$ values (including the vertical asymptote) to obtain $(- \frac{5}{2}, - 1)$ , $open parenthesis minus 2, 0 close parenthesis$ and $open parenthesis minus 1, 1 close parenthesis$ with a vertical asymptote $x = - 3$ . Next, we add $1$ to the $y$ values on the graph and get $(- \frac{5}{2}, 0)$ , $open parenthesis minus 2, 1 close parenthesis$ and $open parenthesis minus 1, 2 close parenthesis$ . If you are experiencing déjà vu, there is a good reason for it but we leave it to the reader to determine the source of this uncanny familiarity. We obtain the graph below. The domain of $f^{- 1}$ is $open parenthesis minus 3, ∞ close parenthesis$ , which matches the range of $f$ , and the range of $f^{- 1}$ is $open parenthesis minus ∞, ∞ close parenthesis$ , which matches the domain of $f$ .
Figure 6.18 $y = j (x) = \log_{2} (x)$
$\overset{}{\to}$
Figure 6.19 $y = f^{- 1} (x) = \log_{2} (x + 3) + 1$
We now verify that $f (x) = 2^{x - 1} - 3$ and $f^{- 1} (x) = \log_{2} (x + 3) + 1$ satisfy the composition requirement for inverses. For all real numbers $x$ ,
$\begin{matrix} (f^{- 1} \circ f) (x) & = & f^{- 1} (f (x)) \\ = & f^{- 1} (2^{x - 1} - 3) \\ = & \log_{2} ([2^{x - 1} - 3] + 3) + 1 \\ = & \log_{2} (2^{x - 1}) + 1 \\ = & (x - 1) + 1 & Since \log_{2} (2^{u}) = u for all real numbers u \\ = & x ✓ \end{matrix}$
For all real numbers $x greater than minus 3$ , we have¹²
$\begin{matrix} (f \circ f^{- 1}) (x) & = & f (f^{- 1} (x)) \\ = & f (\log_{2} (x + 3) + 1) \\ = & 2^{(\log_{2} (x + 3) + 1) - 1} - 3 \\ = & 2^{\log_{2} (x + 3)} - 3 \\ = & (x + 3) - 3 & Since 2^{\log_{2} (u)} = u for all real numbers u > 0 \\ = & x ✓ \end{matrix}$
Last, but certainly not least, we graph $y = f (x)$ and $y = f^{- 1} (x)$ on the same set of axes and see the symmetry about the line $y = x$ .
Figure 6.20

Exercises

In Exercises -, use the property: $b^{a} = c$ if and only if $\log_{b} (c) = a$ from Theorem to rewrite the given equation in the other form. That is, rewrite the exponential equations as logarithmic equations and rewrite the logarithmic equations as exponential equations.

$2 cubed equals 8$
$5^{- 3} = \frac{1}{125}$
$4^{5 / 2} = 32$
${(\frac{1}{3})}^{- 2} = 9$
${(\frac{4}{25})}^{- 1 / 2} = \frac{5}{2}$
$10^{- 3} = 0.001$
$e^{0} = 1$
$log sub 5 ⁡ open parenthesis 25 close parenthesis equals 2$
$\log_{25} (5) = \frac{1}{2}$
$\log_{3} (\frac{1}{81}) = - 4$
$\log_{\frac{4}{3}} (\frac{3}{4}) = - 1$
$log ⁡ open parenthesis 100 close parenthesis equals 2$
$\log (0.1) = - 1$
$ln ⁡ open parenthesis e close parenthesis equals 1$
$\ln (\frac{1}{\sqrt{e}}) = - \frac{1}{2}$
$log sub 3 ⁡ open parenthesis 27 close parenthesis$
$log sub 6 ⁡ open parenthesis 216 close parenthesis$
$log sub 2 ⁡ open parenthesis 32 close parenthesis$
$\log_{6} (\frac{1}{36})$
$log sub 8 ⁡ open parenthesis 4 close parenthesis$
$log sub 36 ⁡ open parenthesis 216 close parenthesis$
$\log_{\frac{1}{5}} (625)$
$\log_{\frac{1}{6}} (216)$
$log sub 36 ⁡ open parenthesis 36 close parenthesis$
$\log (\frac{1}{1000000})$
$log ⁡ open parenthesis 0.01 close parenthesis$
$\ln (e^{3})$
$log sub 4 ⁡ open parenthesis 8 close parenthesis$
$log sub 6 ⁡ open parenthesis 1 close parenthesis$
$log sub 13 ⁡ open parenthesis the square root of 13 close parenthesis$
$log sub 36 ⁡ open parenthesis the 4th root of 36 close parenthesis$
$7 to the power log sub 7 ⁡ open parenthesis 3 close parenthesis$
$36 to the power log sub 36 ⁡ open parenthesis 216 close parenthesis$
$log sub 36 ⁡ open parenthesis 36 to the power 216 close parenthesis$
$ln ⁡ open parenthesis e to the power 5 close parenthesis$
$\log (\sqrt[9]{10^{11}})$
$\log (\sqrt[3]{10^{5}})$
$\ln (\frac{1}{\sqrt{e}})$
$log sub 5 ⁡ open parenthesis 3 to the power log sub 3 ⁡ open parenthesis 5 close parenthesis close parenthesis$
$log ⁡ open parenthesis e to the power ln ⁡ open parenthesis 100 close parenthesis close parenthesis$
$\log_{2} (3^{- \log_{3} (2)})$
$ln ⁡ open parenthesis 42 to the power 6 ⁢ log ⁡ open parenthesis 1 close parenthesis close parenthesis$
$f (x) = \ln (x^{2} + 1)$
$f (x) = \log_{7} (4 x + 8)$
$f (x) = \ln (4 x - 20)$
$f (x) = \log (x^{2} + 9 x + 18)$
$f (x) = \log (\frac{x + 2}{x^{2} - 1})$
$f (x) = \log (\frac{x^{2} + 9 x + 18}{4 x - 20})$
$f (x) = \ln (7 - x) + \ln (x - 4)$
$f (x) = \ln (4 x - 20) + \ln (x^{2} + 9 x + 18)$
$f (x) = \log (x^{2} + x + 1)$
$f (x) = \sqrt[4]{\log_{4} (x)}$
$f (x) = \log_{9} (| x + 3 | - 4)$
$f (x) = \ln (\sqrt{x - 4} - 3)$
$f (x) = \frac{1}{3 - \log_{5} (x)}$
$f (x) = \frac{\sqrt{- 1 - x}}{\log_{\frac{1}{2}} (x)}$
$f (x) = \ln (- 2 x^{3} - x^{2} + 13 x - 6)$
$f (x) = 2^{x}$ , $g (x) = 2^{x} - 1$
$f (x) = {(\frac{1}{3})}^{x}$ , $g (x) = {(\frac{1}{3})}^{x - 1}$
$f (x) = 3^{x}$ , $g (x) = 3^{- x} + 2$
$f (x) = 10^{x}$ , $g (x) = 10^{\frac{x + 1}{2}} - 20$
$f (x) = e^{x}$ , $g (x) = 8 - e^{- x}$
$f (x) = e^{x}$ , $g (x) = 10 e^{- 0.1 x}$
$f ⁢ open parenthesis x close parenthesis equals log sub 2 ⁡ open parenthesis x close parenthesis$ , $g (x) = \log_{2} (x + 1)$
$f (x) = \log_{\frac{1}{3}} (x)$ , $g (x) = \log_{\frac{1}{3}} (x) + 1$
$f ⁢ open parenthesis x close parenthesis equals log sub 3 ⁡ open parenthesis x close parenthesis$ , $g (x) = - \log_{3} (x - 2)$
$f (x) = \log (x)$ , $g (x) = 2 \log (x + 20) - 1$
$f (x) = \ln (x)$ , $g (x) = - \ln (8 - x)$
$f (x) = \ln (x)$ , $g (x) = - 10 \ln (\frac{x}{10})$
Verify that each function in Exercises - is the inverse of the corresponding function in Exercises -. (Match up # and #, and so on.)
$f (x) = 3^{x + 2} - 4$
$f (x) = \log_{4} (x - 1)$
$f (x) = - 2^{- x} + 1$
$f (x) = 5 \log (x) - 2$
Earthquakes are complicated events and it is not our intent to provide a complete discussion of the science involved in them. Instead, we refer the interested reader to a solid course in Geology¹³ or the U.S. Geological Survey's Earthquake Hazards Program found here and present only a simplified version of the Richter scale . The Richter scale measures the magnitude of an earthquake by comparing the amplitude of the seismic waves of the given earthquake to those of a “magnitude 0 event”, which was chosen to be a seismograph reading of $0.001$ millimeters recorded on a seismometer 100 kilometers from the earthquake's epicenter. Specifically, the magnitude of an earthquake is given by
$M (x) = \log (\frac{x}{0.001})$
where $x$ is the seismograph reading in millimeters of the earthquake recorded 100 kilometers from the epicenter.
1. Show that $M ⁢ open parenthesis 0.001 close parenthesis equals 0$ .
2. Compute $M ⁢ open parenthesis 80, 000 close parenthesis$ .
3. Show that an earthquake which registered 6.7 on the Richter scale had a seismograph reading ten times larger than one which measured 5.7.
4. Find two news stories about recent earthquakes which give their magnitudes on the Richter scale. How many times larger was the seismograph reading of the earthquake with larger magnitude?
While the decibel scale can be used in many disciplines,¹⁴ we shall restrict our attention to its use in acoustics, specifically its use in measuring the intensity level of sound.¹⁵ The Sound Intensity Level $L$ (measured in decibels) of a sound intensity $I$ (measured in watts per square meter) is given by
$L (I) = 10 \log (\frac{I}{10^{- 12}}) .$
Like the Richter scale, this scale compares $I$ to baseline: $10^{- 12} \frac{W}{m^{2}}$ is the threshold of human hearing.
1. Compute $L (10^{- 6})$ .
2. Damage to your hearing can start with short term exposure to sound levels around 115 decibels. What intensity $I$ is needed to produce this level?
3. Compute $L ⁢ open parenthesis 1 close parenthesis$ . How does this compare with the threshold of pain which is around 140 decibels?
The pH of a solution is a measure of its acidity or alkalinity. Specifically, $pH = - \log [H^{+}]$ where $[H^{+}]$ is the hydrogen ion concentration in moles per liter. A solution with a pH less than 7 is an acid, one with a pH greater than 7 is a base (alkaline) and a pH of 7 is regarded as neutral.
1. The hydrogen ion concentration of pure water is $[H^{+}] = 10^{- 7}$ . Find its pH.
2. Find the pH of a solution with $[H^{+}] = 6.3 \times 10^{- 13}$ .
3. The pH of gastric acid (the acid in your stomach) is about $0.7$ . What is the corresponding hydrogen ion concentration?
Show that $log sub b ⁡ 1 equals 0$ and $log sub b ⁡ b equals 1$ for every $b greater than 0, b not equal to 1$ .
(Crazy bonus question) Without using your calculator, determine which is larger: $e^{π}$ or $π^{e}$ .

In Exercises -, evaluate the expression.

In Exercises -, find the domain of the function.

In Exercises -, sketch the graph of $y = g (x)$ by starting with the graph of $y = f (x)$ and using transformations. Track at least three points of your choice and the horizontal asymptote through the transformations. State the domain and range of $g$ .

In Exercises -, sketch the graph of $y = g (x)$ by starting with the graph of $y = f (x)$ and using transformations. Track at least three points of your choice and the vertical asymptote through the transformations. State the domain and range of $g$ .

In Exercises -, find the inverse of the function from the `procedural perspective' discussed in Example Example 5 and graph the function and its inverse on the same set of axes.

(Logarithmic Scales) In Exercises -, we introduce three widely used measurement scales which involve common logarithms: the Richter scale, the decibel scale and the pH scale. The computations involved in all three scales are nearly identical so pay attention to the subtle differences.

Answers

$log sub 2 ⁡ open parenthesis 8 close parenthesis equals 3$
$\log_{5} (\frac{1}{125}) = - 3$
$\log_{4} (32) = \frac{5}{2}$
$\log_{\frac{1}{3}} (9) = - 2$
$\log_{\frac{4}{25}} (\frac{5}{2}) = - \frac{1}{2}$
$\log (0.001) = - 3$
$ln ⁡ open parenthesis 1 close parenthesis equals 0$
$5 squared equals 25$
${(25)}^{\frac{1}{2}} = 5$
$3^{- 4} = \frac{1}{81}$
${(\frac{4}{3})}^{- 1} = \frac{3}{4}$
$10 squared equals 100$
$10^{- 1} = 0.1$
$e^{1} = e$
$e^{- \frac{1}{2}} = \frac{1}{\sqrt{e}}$
$log sub 3 ⁡ open parenthesis 27 close parenthesis equals 3$
$log sub 6 ⁡ open parenthesis 216 close parenthesis equals 3$
$log sub 2 ⁡ open parenthesis 32 close parenthesis equals 5$
$\log_{6} (\frac{1}{36}) = - 2$
$\log_{8} (4) = \frac{2}{3}$
$\log_{36} (216) = \frac{3}{2}$
$\log_{\frac{1}{5}} (625) = - 4$
$\log_{\frac{1}{6}} (216) = - 3$
$log sub 36 ⁡ open parenthesis 36 close parenthesis equals 1$
$\log \frac{1}{1000000} = - 6$
$\log (0.01) = - 2$
$ln ⁡ open parenthesis e cubed close parenthesis equals 3$
$\log_{4} (8) = \frac{3}{2}$
$log sub 6 ⁡ open parenthesis 1 close parenthesis equals 0$
$\log_{13} (\sqrt{13}) = \frac{1}{2}$
$\log_{36} (\sqrt[4]{36}) = \frac{1}{4}$
$7^{\log_{7} (3)} = 3$
$36^{\log_{36} (216)} = 216$
$\log_{36} (36^{216}) = 216$
$\ln (e^{5}) = 5$
$\log (\sqrt[9]{10^{11}}) = \frac{11}{9}$
$\log (\sqrt[3]{10^{5}}) = \frac{5}{3}$
$\ln (\frac{1}{\sqrt{e}}) = - \frac{1}{2}$
$\log_{5} (3^{\log_{3} 5}) = 1$
$\log (e^{\ln (100)}) = 2$
$\log_{2} (3^{- \log_{3} (2)}) = - 1$
$\ln (42^{6 \log (1)}) = 0$
$open parenthesis minus ∞, ∞ close parenthesis$
$open parenthesis minus 2, ∞ close parenthesis$
$open parenthesis 5, ∞ close parenthesis$
$(- \infty, - 6) \cup (- 3, \infty)$
$(- 2, - 1) \cup (1, \infty)$
$(- 6, - 3) \cup (5, \infty)$
$open parenthesis 4, 7 close parenthesis$
$open parenthesis 5, ∞ close parenthesis$
$open parenthesis minus ∞, ∞ close parenthesis$
$open bracket 1, ∞ close parenthesis$
$(- \infty, - 7) \cup (1, \infty)$
$open parenthesis 13, ∞ close parenthesis$
$open parenthesis 0, 125 close parenthesis union open parenthesis 125, ∞ close parenthesis$
No domain
$(- \infty, - 3) \cup (\frac{1}{2}, 2)$
Domain of $g$ : $open parenthesis minus ∞, ∞ close parenthesis$ Range of $g$ : $open parenthesis minus 1, ∞ close parenthesis$
Figure 6.21 $y = g (x) = 2^{x} - 1$
Domain of $g$ : $open parenthesis minus ∞, ∞ close parenthesis$ Range of $g$ : $open parenthesis 0, ∞ close parenthesis$
Figure 6.22 $y = g (x) = {(\frac{1}{3})}^{x - 1}$
Domain of $g$ : $open parenthesis minus ∞, ∞ close parenthesis$ Range of $g$ : $open parenthesis 2, ∞ close parenthesis$
Figure 6.23 $y = g (x) = 3^{- x} + 2$
Domain of $g$ : $open parenthesis minus ∞, ∞ close parenthesis$ Range of $g$ : $open parenthesis minus 20, ∞ close parenthesis$
Figure 6.24 $y = g (x) = 10^{\frac{x + 1}{2}} - 20$
Domain of $g$ : $open parenthesis minus ∞, ∞ close parenthesis$ Range of $g$ : $open parenthesis minus ∞, 8 close parenthesis$
Figure 6.25 $y = g (x) = 8 - e^{- x}$
Domain of $g$ : $open parenthesis minus ∞, ∞ close parenthesis$ Range of $g$ : $open parenthesis 0, ∞ close parenthesis$
Figure 6.26 $y = g (x) = 10 e^{- 0.1 x}$
Domain of $g$ : $open parenthesis minus 1, ∞ close parenthesis$ Range of $g$ : $open parenthesis minus ∞, ∞ close parenthesis$
Figure 6.27 $y = g (x) = \log_{2} (x + 1)$
Domain of $g$ : $open parenthesis 0, ∞ close parenthesis$ Range of $g$ : $open parenthesis minus ∞, ∞ close parenthesis$
Figure 6.28 $y = g (x) = \log_{\frac{1}{3}} (x) + 1$
Domain of $g$ : $open parenthesis 2, ∞ close parenthesis$ Range of $g$ : $open parenthesis minus ∞, ∞ close parenthesis$
Figure 6.29 $y = g (x) = - \log_{3} (x - 2)$
Domain of $g$ : $open parenthesis minus 20, ∞ close parenthesis$ Range of $g$ : $open parenthesis minus ∞, ∞ close parenthesis$
Figure 6.30 $y = g (x) = 2 \log (x + 20) - 1$
Domain of $g$ : $open parenthesis minus ∞, 8 close parenthesis$ Range of $g$ : $open parenthesis minus ∞, ∞ close parenthesis$
Figure 6.31 $y = g (x) = - \ln (8 - x)$
Domain of $g$ : $open parenthesis 0, ∞ close parenthesis$ Range of $g$ : $open parenthesis minus ∞, ∞ close parenthesis$
Figure 6.32 $y = g (x) = - 10 \ln (\frac{x}{10})$
$f (x) = 3^{x + 2} - 4$ $f^{- 1} (x) = \log_{3} (x + 4) - 2$
Figure 6.33
$f (x) = \log_{4} (x - 1)$ $f^{- 1} (x) = 4^{x} + 1$
Figure 6.34
$f (x) = - 2^{- x} + 1$ $f^{- 1} (x) = - \log_{2} (1 - x)$
Figure 6.35
$f (x) = 5 \log (x) - 2$ $f^{- 1} (x) = 10^{\frac{x + 2}{5}}$
Figure 6.36
1. $M (0.001) = \log (\frac{0.001}{0.001}) = \log (1) = 0$ .
2. $M (80, 000) = \log (\frac{80, 000}{0.001}) = \log (80, 000, 000) \approx 7.9$ .
1. $L (10^{- 6}) = 60$ decibels.
2. $I = 10^{- .5} \approx 0.316$ watts per square meter.
3. Since $L ⁢ open parenthesis 1 close parenthesis equals 120$ decibels and $L ⁢ open parenthesis 100 close parenthesis equals 140$ decibels, a sound with intensity level 140 decibels has an intensity 100 times greater than a sound with intensity level 120 decibels.
1. The pH of pure water is 7.
2. If $[H^{+}] = 6.3 \times 10^{- 13}$ then the solution has a pH of 12.2.
3. $[H^{+}] = 10^{- 0.7} \approx .1995$ moles per liter.

Take a class in Differential Equations and you'll see why. ↩
Recall that this means there are no holes or other kinds of breaks in the graph. ↩
You can actually prove this by considering the polynomial $p (x) = x^{2} - 3$ and showing it has no rational zeros by applying Theorem . ↩
This is where Calculus and continuity come into play. ↩
Want more information? Look up “convergent sequences” on the Internet. ↩
Meaning, graph some more examples on your own. ↩
The proof of which, like many things discussed in the text, requires Calculus. ↩
Recall that this means the graph of $f$ has no sharp turns or corners. ↩
We will discuss this in greater detail in Section . ↩
It is worth a moment of your time to think your way through why $117^{\log_{117} (6)} = 6$ . By definition, $log sub 117 ⁡ open parenthesis 6 close parenthesis$ is the exponent we put on $117$ to get $6$ . What are we doing with this exponent? We are putting it on $117$ . By definition we get $6$ . In other words, the exponential function $f (x) = 117^{x}$ undoes the logarithmic function $g ⁢ open parenthesis x close parenthesis equals log sub 117 ⁡ open parenthesis x close parenthesis$ . ↩
See page if you've forgotten what this term means. ↩
Pay attention - can you spot in which step below we need $x greater than minus 3$ ? ↩
Rock-solid, perhaps? ↩
See this webpage for more information. ↩
As of the writing of this exercise, the Wikipedia page given here states that it may not meet the “general notability guideline” nor does it cite any references or sources. I find this odd because it is this very usage of the decibel scale which shows up in every College Algebra book I have read. Perhaps those other books have been wrong all along and we're just blindly following tradition. ↩

Adapted from Precalculus, 3rd corrected edition, by Carl Stitz and Jeff Zeager (stitz-zeager.com), licensed under CC BY-NC-SA 3.0. Changes were made: reformatted as an accessible XYZ web edition. License: CC-BY-NC-SA-3.0.