6.1: Introduction to Exponential and Logarithmic Functions

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Corequisite Course Topics

• Algebra
  • Systems of Equations
    • Solving
      • via the Substitution Method

If you are consistently struggling with these skills, or if you would feel more comfortable reviewing these skills in a structured environment, you might be better served in a course with support (if you are not already enrolled in such a course).

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Learning Objectives

• Graph and analyze exponential and logarithmic functions involving algebraic transformations (shifts, scale factors, reflections, and absolute value) in the rectangular coordinate system.
• Determine the model for exponential applications.

Exponential Functions

Of all of the functions we study in this text, exponential and logarithmic functions are the ones that 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 $x^{2 / 3}$, 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 $f(x)=2^{x}$. We make a table of values, plot the points, and connect the dots in a pleasing fashion.

$$\begin{array}{|r||r|r|} \hline x & f(x) & (x, f(x)) \\[16pt] \hline-3 & 2^{-3}=\dfrac{1}{8} & \left(-3, \dfrac{1}{8}\right) \\[16pt] \hline-2 & 2^{-2}=\dfrac{1}{4} & \left(-2, \dfrac{1}{4}\right) \\[16pt] \hline-1 & 2^{-1}=\dfrac{1}{2} & \left(-1, \dfrac{1}{2}\right) \\[16pt] \hline 0 & 2^{0}=1 & (0,1) \\[16pt] \hline 1 & 2^{1}=2 & (1,2) \\[16pt] \hline 2 & 2^{2}=4 & (2,4) \\[16pt] \hline 3 & 2^{3}=8 & (3,8) \\[16pt] \hline \end{array}\nonumber$$

A few remarks about the graph $f(x)=2^{x}$ which we have constructed are in order. As $x \to-\infty$ 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,

$$\text{as } x \to-\infty, \, 2^{x} \approx \dfrac{1}{\text{very big }(+)} \approx \text {very small }(+) \to 0^+. \nonumber$$ This is represented graphically using the $x$-axis (the line $y = 0$) as a horizontal asymptote.

On the flip side, as $x \to \infty$, we find $f(100)=2^{100}$, $f(1000)=2^{1000}$, and so on. Thus, $2^{x} \to \infty$. As a result, our graph suggests the range of $f$ is $(0, \infty)$. The graph of $f$ passes the Horizontal Line Test, which means $f$ is one-to-one and, as such, is 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 $\sqrt{3} = 1.73205 \ldots$ is irrational ³ and as such, its decimal representation neither repeats nor terminates. We can, however, approximate $\sqrt{3}$ by terminating decimals, and it stands to reason⁴ we can use these to approximate $2^{\sqrt{3}}$. For example, if we approximate $\sqrt{3}$ by $1.73$, we can approximate $2^{\sqrt{3}}$ using

$$2^{\sqrt{3}} \approx 2^{1.73} = 2^{\frac{173}{100}} = \sqrt[100]{2^{173}}. \nonumber$$ 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 $\sqrt{3}$ 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 $b<0$. 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 $b \geq 0$.

What about $b = 0$?

The function $f(x)=0^{x}$ is undefined for $x \leq 0$ because we cannot divide by $0$ and $0^{0}$ is an indeterminate form. For $x>0$, $0^{x}=0$ so the function $f(x)=0^{x}$ is the same as the function $f(x)=0$ (again, for $x>0$). We know everything about this function and exclude it from our investigations. The only other base we exclude is $b = 1$ 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.

Definition: Exponential Function

A function of the form $f(x)=b^{x}$ where $b$ is a fixed real number, $b>0$, and $b \neq 1$, is called a base $b$ exponential function.

We leave it to the reader to verify⁶ that if $b>1$, then the exponential function $f(x)=b^{x}$ will share the same basic shape and characteristics as $f(x)=2^{x}$.

What if $0<b<1$?

Consider $g(x)=\left(\frac{1}{2}\right)^{x}$. We could certainly build a table of values and connect the points, or we could take a step back and note that $g(x)=\left(\frac{1}{2}\right)^{x}=\left(2^{-1}\right)^{x}=2^{-x}=f(-x)$. Thinking back to Section 1.7, the graph of $f(-x)$ is obtained from the graph of $f(x)$ by reflecting it across the $y$-axis. We get

We see that the domain and range of $g$ match that of $f$, namely $(-\infty, \infty)$ and $(0, \infty)$, respectively. Like $f$, $g$ is also one-to-one. Whereas $f$ is always increasing, $g$ is always decreasing. As a result, as $x \to -\infty$, $g(x) \to \infty$, and on the flip side, as $x \to \infty$, $g(x) \to 0^{+}$. It shouldn't be too surprising that for all choices of the base $0<b<1$, the graph $y=b^{x}$ 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

Let $f(x)=b^{x}$.

• The domain of $f$ is $(-\infty, \infty)$ and the range of $f$ is $(0, \infty)$.
• $(0, 1)$ is on the graph of $f$ and $y = 0$ is a horizontal asymptote to the graph of $f$.
• $f$ is one-to-one, continuous and smooth.⁸
• If $b>1$:
  • $f$ is always increasing
  • As $x \to-\infty$, $f(x) \to 0^{+}$
  • As $x \to \infty, f(x) \to \infty$
  • The graph of $f$ resembles:

    

• If $0<b<1$:
  • $f$ is always decreasing
  • As $x \to-\infty, f(x) \to \infty$
  • As $x \to-\infty, f(x) \to \infty$
  • As $x \to \infty, f(x) \to 0^{+}$
  • The graph of $f$ resembles:

    

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, $e \approx 2.718$, called the natural base. We will more formally discuss the origins of this number in Section 6.5. For now, it is enough to know that since $e>1$, $f(x)=e^{x}$, is an increasing exponential function.

Finding Equations of Exponential Functions

Sometimes, we need to find the equation for an exponential function going through two given points. To do so, we employ a method developed in Elementary Algebra. Specifically, we build and solve a system of equations.

Example $\PageIndex{1}$

Find the equation of the exponential function of the form $f(x) = a b^x$ whose graph contains the points $(3, 54)$ and $\left(6, \frac{729}{256} \right)$.

Solution

The key here is that we want an exponential function of the form $f(x) = a b^x$. Without that instruction, we would have an infinite number of exponential functions going through the given points. Since both points lie on the graph of $f$, we know that $$54 = a b^3 \nonumber$$ and $$\dfrac{729}{256} = ab^6. \nonumber$$ These two equations form a system of two (nonlinear) equations in two unknowns.

We can solve the first equation for $a$ to get $a = \frac{54}{b^3}$. Using the Substitution Method from Elementary Algebra, we substitute this value of $a$ into the second equation to find $$\dfrac{729}{256} = \dfrac{54}{b^3} \cdot b^6 \implies \dfrac{729}{256} = \dfrac{54b^6}{b^3}. \nonumber$$ Using the Laws of Exponents, we get $$\dfrac{729}{256} = 54b^3 \implies \dfrac{27}{512} = b^3 \implies \dfrac{3}{8} = b. \nonumber$$ Continuing with the Substitution Method, we allow $b = \frac{3}{8}$ in either of the equations above. Let's choose the one with the smaller exponent. $$54 = a \left( \dfrac{3}{8} \right)^3 \implies 54 = \dfrac{27}{512} a \implies 1024 = a. \nonumber$$ Thus, the equation of the exponential is $$f(x) = 1024 \left(\dfrac{3}{8}\right)^x. \nonumber$$

Applications Involving Exponential Functions

The following examples give us an idea of how exponential functions are used in the wild.

Example $\PageIndex{2}$

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

1. Find and interpret $V (0)$.
2. Sketch the graph of $y = V (x)$ using transformations.
3. Find and interpret the horizontal asymptote of the graph you found in part b.

Solutions

1. To find $V (0)$, we replace $x$ with $0$ to obtain $V(0)=25\left(\frac{4}{5}\right)^{0}=25$. Since $x$ represents the car's age in years, $x = 0$ corresponds to the car being brand new. Since $V (x)$ is measured in thousands of dollars, $V (0) = 25$ corresponds to a value of $25,000. Putting it all together, we interpret $V (0) = 25$ to mean the car's purchase price was $25,000.
2. To graph $y=25\left(\frac{4}{5}\right)^{x}$, we start with the basic exponential function $f(x)=\left(\frac{4}{5}\right)^{x}$. 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 $(0, 1)$ and two other points, $\left(-1, \frac{5}{4}\right)$ and $\left(1, \frac{4}{5}\right)$, and label the horizontal asymptote $y=0$. To obtain $V(x)=25\left(\frac{4}{5}\right)^{x}, x \geq 0$, we multiply the output from $f$ by $25$, in other words, $V (x) = 25f(x)$. 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 $\left(-1, \frac{125}{4}\right)$, $(0, 25)$ and $(1, 20)$. The horizontal asymptote remains $y = 0$. Finally, we restrict the domain to $[0, \infty)$ to fit with the applied domain given to us. We have the result below.

   

3. We see from the graph of $V$ that its horizontal asymptote is $y = 0$. (We leave it to the 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 6.5. For now, we present another typical 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 just a basic exponential function that has been modified by a few transformations from Section 1.7.

Example $\PageIndex{3}$

According to Newton’s Law of Cooling⁹ the temperature of coffee $T$ (in degrees Fahrenheit) $t$ minutes after it is served can be modeled by $T(t)=70+90 e^{-0.1 t}$.

1. Find and interpret $T(0)$.
2. Sketch the graph of $y = T(t)$ using transformations.
3. Find and interpret the horizontal asymptote of the graph.

Solutions

1. To find $T(0)$, 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 $160^{\circ}$F.
2. To graph $y = T(t)$ using transformations, we start with the basic function, $f(t)=e^{t}$. As we have already remarked, $e \approx 2.718>1$ so the graph of $f$ is an increasing exponential with $y$-intercept $(0, 1)$ and horizontal asymptote $y = 0$. The points $\left(-1, e^{-1}\right) \approx(-1,0.37)$ and $(1, e) \approx(1,2.72)$ are also on the graph. Since the formula $T(t)$ looks rather complicated, we rewrite $T(t)$ in the form $$T(t)=70+90 e^{-0.1 t}=90 e^{-0.1 t}+70=90 f(-0.1 t)+70\nonumber$$ and use this to track the change to our three points and the horizontal asymptote. Multiplication of the input of $f$, $t$, by $−0.1$ 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 $−0.1$. (which amounts to multiplying them by $−10$) to obtain $\left(10, e^{-1}\right)$, $(0, 1)$, and $(−10, e)$. Since none of these changes affected the $y$ values, the horizontal asymptote remains $y = 0$. Next, the output from $f$ is multiplied by $90$. This results in a vertical stretch by a factor of $90$. We multiply the $y$-coordinates by $90$ to obtain $\left(10,90 e^{-1}\right)$, $(0, 90)$, $(−10, 90e)$. We also multiply the $y$ value of the horizontal asymptote $y = 0$ by $90$, and it remains $y = 0$. Finally, we add $70$ to all of the $y$-coordinates, which shifts the graph upwards to obtain $\left(10,90 e^{-1}+70\right) \approx(10,103.11)$, $(0, 160)$, and $(-10,90 e+70) \approx(-10,314.64)$. Adding $70$ to the horizontal asymptote also shifts it upwards to $y = 70$. 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 $[0, \infty)$. The result is below.

   

3. From the graph, the horizontal asymptote is $y = 70$. It is worth a moment to see how this happens analytically and review some of the "number sense" developed in Chapter 4. As $t \to \infty$, We get $T(t)=70+90 e^{-0.1 t} \approx 70+90 e^{\text {very big }(-)}$. Since $e > 1$, $$e^{\text {very big }(-)}=\dfrac{1}{e^{\text {very big }(+)}} \approx \dfrac{1}{\text {very big }(+)} \approx \text{very small }(+).\nonumber$$ The larger $t$ becomes, the smaller $e^{-0.1 t}$ becomes, so the term $90 e^{-0.1 t} \approx \text { very small }(+)$. Hence, $T(t) \approx 70 + \text { very small }(+)$ which means the graph is approaching the horizontal line $y = 70$ from above. This means that as time goes by, the temperature of the coffee is cooling to $70^{\circ} \mathrm{F}$, presumably room temperature.

Logarithmic Functions

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 focus on these inverses, the logarithmic functions called "logs" for short.

Definition: Logarithmic Function

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, $b = 10$, and the natural base, $b = e$.

Definition: Common and Natural Logarithms

The common logarithm of a real number $x$ is $\log _{10}(x)$ 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 logarithms are defined as the inverses of exponential functions, we can use the properties of inverse functions, their uniqueness, and their graphs 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 $(0, \infty)$, and that the range of a log function is the domain of an exponential function, namely $(-\infty, \infty)$. 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 $(0, 1)$ on the graph of $f$ corresponds to an $x$-intercept of $(1, 0)$ on the graph of $f^{-1}$. The horizontal asymptotes $y = 0$ on the graphs of the exponential functions become vertical asymptotes $x = 0$ on the log graphs.

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, $\log _{2}(8)=3$. More generally, $\log _{2}(x)$ is the exponent you put on $2$ to get $x$. Thus, $\log _{2}(16)=4$, 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 $(0, \infty)$ and the range of $f$ is $(-\infty, \infty)$.
• $(1, 0)$ is on the graph of $f$ and $x = 0$ 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 $c$.
• $\log _{b}\left(b^{x}\right)=x$ for all $x$ and $b^{\log _{b}(x)}=x$ for all $x > 0$
• If $b > 1$:
  • $f$ is always increasing
  • As $x \to 0^{+}, f(x) \to-\infty$
  • As $x \to \infty, f(x) \to \infty$
  • The graph of $f$ resembles:

    

• If $0 < b < 1$:
  • $f$ is always decreasing
  • As $x \to 0^{+}, f(x) \to \infty$
  • As $x \to \infty, f(x) \to-\infty$
  • The graph of $f$ resembles:

    

As we have mentioned, this theorem is a consequence of the properties of inverse functions and the uniqueness of inverse functions and their graphs. However, it is worth the reader's time to understand this theorem from an exponential perspective. For instance, we know that the domain of $g(x)=\log _{2}(x)$ is $(0, \infty)$. Why? Because the range of $f(x)=2^{x}$ is $(0, \infty)$. In a way, this says everything, but at the same time, it doesn't. For example, if we try to find $\log _{2}(-1)$, we are trying to find the exponent we put on $2$ to give us $-1$. 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 the same thing as saying "the domain of $g(x)=\log _{2}(x)$ is $(0, \infty)$ because the range of $f(x)=2^{x}$ is $(0, \infty)$," we feel it is in a student's best interest to understand the statements in the previous theorem at this level instead of just merely memorizing the facts.

Example $\PageIndex{4}$

Simplify the following.

1. $\log _{3}(81)$
2. $\log _{2}\left(\frac{1}{8}\right)$
3. $\log _{\sqrt{5}}(25)$
4. $\ln \left(\sqrt[3]{e^{2}}\right)$
5. $\log (0.001)$
6. $2^{\log _{2}(8)}$
7. $117^{-\log _{117}(6)}$

Solutions

1. The number $\log _{3}(81)$ 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 $\log _{3}(81)=4$.
2. To find $\log _{2}\left(\frac{1}{8}\right)$, we need rewrite $\frac{1}{8}$ as a power of $2$. We find $\frac{1}{8}=\frac{1}{2^{3}}=2^{-3}$, so $\log _{2}\left(\frac{1}{8}\right)=-3$.
3. To determine $\log _{\sqrt{5}}(25)$, we need to express $25$ as a power of $\sqrt{5}$. We know $25=5^{2}$, and $5=(\sqrt{5})^{2}$, so we have $25=\left((\sqrt{5})^{2}\right)^{2}=(\sqrt{5})^{4}$. We get $\log _{\sqrt{5}}(25)=4$.
4. First, recall that the not $\ln \left(\sqrt[3]{e^{2}}\right)$ means $\log _{e}\left(\sqrt[3]{e^{2}}\right)$, 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 $\ln \left(\sqrt[3]{e^{2}}\right)=\ln \left(e^{2 / 3}\right)=\frac{2}{3}$.
5. Rewriting $\log (0.001)$ as $\log _{10}(0.001)$, 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, $\log (0.001)=\log \left(10^{-3}\right)=-3$.
6. We can use Theorem $\PageIndex{2}$ directly to so simplify $2^{\log _{2}(8)}=8$. We can also understand this problem by first finding $\log _{2}(8)$. By definition, $\log _{2}(8)$ is the exponent we put on $2$ to get $8$. Since $8=2^{3}$, we have $\log _{2}(8)=3$. We now substitute to find $2^{\log _{2}(8)}=2^{3}=8$.
7. $117^{-\log _{117}(6)}=\frac{1}{117^{\log _{117}(6)}}=\frac{1}{6}$

Up until this point, restrictions on the domains of functions came from avoiding division by zero and keeping negative numbers from beneath even-indexed radicals. With the introduction of logarithms, we now have another restriction. Since the domain of $f(x)=\log _{b}(x)$ is $(0, \infty)$, the argument of the log must be strictly positive.

Example $\PageIndex{5}$

Find the domain of the following functions. Check your answers using graphing technology.

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

Solutions

1. We set $3-x>0$ to obtain $x<3$, or $(-\infty, 3)$. The graph from the calculator below verifies this. Note that we could have graphed $f$ using transformations by rewriting $f(x)=2 \log _{10}(-x+3)-1$, and recognizing the main function involved is $y=h(x)=\log _{10}(x)$. We select three points to track, $\left(\frac{1}{10},-1\right),(1,0) \text { and }(10,1)$, along with the vertical asymptote $x = 0$. Since $f(x)=2 h(-x+3)-1$, we first subtract $3$ from the $x$-coordinates (shifting the graph left $3$ units), then divide (multiply) by the $x$-coordinates by $-1$ (causing a reflection across the $y$-axis). These transformations apply to the vertical asymptote $x = 0$. Subtracting $3$ gives us $x = −3$ as our asymptote, then multiplying by $-1$ gives us the vertical asymptote $x = 3$. 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.
2. 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 $x = 1$ and $r(x) = 0$ when $x = 0$. Choosing some test values, we generate the sign diagram below.

   

   We find $\frac{x}{x-1}>0$ on $(-\infty, 0) \cup(1, \infty)$ 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 1.7 transformation 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 $x \to \pm \infty$, $g(x) \to 0$. We can verify this analytically. Using results from Chapter 4 and continuity, we know that as $x \to \pm \infty$, $\frac{x}{x-1} \approx 1$. Hence, it makes sense that $g(x)=\ln \left(\frac{x}{x-1}\right) \approx \ln (1)=0$.

   

While logarithms have some interesting applications that 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 the material in the section.

Example $\PageIndex{6}$

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

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

Solutions

1. If we identify $g(x)=2^{x}$, we see $f(x)=g(x-1)-3$. We pick the points $\left(-1, \frac{1}{2}\right),(0,1)$ and $(1,2)$ on the graph of $g$ along with the horizontal asymptote $y = 0$ to track through the transformations. 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 $\left(0, \frac{1}{2}\right),(1,1) \text { and }(2,2)$. The horizontal asymptote remains $y = 0$. Next, we subtract $3$ from the $y$-coordinates, shifting the graph down $3$ units. We get the points $\left(0,-\frac{5}{2}\right),(1,-2) \text { and }(2,-1)$ 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 $(-\infty, \infty)$, but that the range of $f$ is $(-3, \infty)$.

   

2. 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
   i. subtract $1$
   ii. put as an exponent on $2$
   iii. 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, $\log _{2}(x)$ undoes exponentiation by $2$. Hence, $f^{-1}$ should
   i. add $3$
   ii. take the logarithm base $2$
   iii. add $1$
   In symbols, $f^{-1}(x)=\log _{2}(x+3)+1$.
3. To graph $f^{-1}(x)=\log _{2}(x+3)+1$ using transformations, we start with $j(x)=\log _{2}(x)$. We track the points $\left(\frac{1}{2},-1\right)$, $(1,0)$, and $(2,1)$ on the graph of $j$ along with the vertical asymptote $x = 0$ through the transformations. Since $f^{-1}(x)=j(x+3)+1$, we first subtract $3$ from each of the $x$ values (including the vertical asymptote) to obtain $\left(-\frac{5}{2},-1\right)$, (−2, 0) and (−1, 1) with a vertical asymptote $x = −3$. Next, we add $1$ to the $y$ values on the graph and get $\left(-\frac{5}{2}, 0\right)$, $(−2, 1)$, and $(−1, 2)$. If you are experiencing deja vu, there is a good reason for it. Still, 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 $(-3, \infty)$, which matches the range of $f$, and the range of $f^{-1}$ is $(-\infty, \infty)$, which matches the domain of $f$.

   

4. 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 $x$, $$\begin{array}{rclcl} \left(f^{-1} \circ f\right)(x) & = & f^{-1}(f(x)) & & \\ & = & f^{-1}\left(2^{x-1}-3\right) & & \\ & = & \log _{2}\left(\left[2^{x-1}-3\right]+3\right)+1 & & \\ & = & \log _{2}\left(2^{x-1}\right)+1 & & \\ & = & (x-1)+1 & \quad & \left( \text{Since } \log _{2}\left(2^{u}\right)=u \text { for all real numbers } u\right) \\ & = & x & & \checkmark \\ \end{array}\nonumber$$ For all real numbers $x > −3$, we have¹⁰ $$\begin{array}{rclcl} \left(f \circ f^{-1}\right)(x) & = f\left(f^{-1}(x)\right) & & \\ & = & f\left(\log _{2}(x+3)+1\right) & & \\ & = & 2^{\left(\log _{2}(x+3)+1\right)-1}-3 & & \\ & = & 2^{\log _{2}(x+3)}-3 & & \\ & = & (x+3)-3 & \quad & \left(\text{Since } 2^{\log _{2}(u)}=u \text { for all real numbers } u>0\right) \\ & = & x & \checkmark \\ \end{array}\nonumber$$
5. 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 $y=x$.

   

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Footnotes

¹ Take a class in Differential Equations and you'll see why.

² Recall that this means there are no holes or other kinds of breaks in the graph.

³ You can prove this by considering the polynomial $p(x)=x^{2}-3$ and showing it has no rational zeros by applying the Rational Zeros Theorem.

⁴ This is where Calculus and continuity come into play.

⁵ Want more information? Look up "convergent sequences" on the Internet.

⁶ Meaning, graph some more examples on your own.

⁷ The proof of which, like many things discussed in the text, requires Calculus.

⁸ Recall that this means the graph of $f$ has no sharp turns or corners.

⁹ We will discuss this in greater detail in Section 6.5.

¹⁰ Pay attention - can you spot in which step below we need $x>-3$?