Inverses of Exponential Functions - AP PreCalculus Study Guide

The Core Idea: Inverses of Exponential Functions

The fundamental concept of this topic is that for every exponential function, there exists an inverse function that "undoes" its operation. This inverse is known as a logarithmic function. Specifically, an exponential function of the form $f(x)=b^x$ (where the base $b$ is positive and not equal to 1) maps an input (an exponent) to an output (the resulting value). The inverse function, a logarithm, performs the reverse operation: it takes the resulting value as its input and determines the original exponent.

This inverse relationship has profound implications for the properties of these functions. The process of finding an inverse involves swapping the roles of inputs and outputs, which means the domain of the exponential function becomes the range of its logarithmic inverse, and the range of the exponential becomes the domain of the logarithm. Graphically, this relationship is visualized as a perfect reflection across the line $y=x$. Understanding this core idea is crucial for solving exponential equations and analyzing the behavior of both function types.

Key Formulas/Rules

The relationship between an exponential function and its inverse is defined by the following rules, derived directly from the properties of inverse functions.

The Inverse Pair

For an exponential function $f(x)=b^x$, where $b>0$ and $b\ne 1$, its inverse function is the logarithmic function with the same base:

$$f^{-1}(x)={\log }_b(x)$$

The Composition Property

The composition of a function and its inverse results in the identity function, $g(x)=x$. This means that applying an exponential function and its corresponding logarithmic function in succession will return the original input value.

1. Logarithm of an Exponential: Applying the logarithmic function after the exponential function.

   $$f^{-1}(f(x))={\log }_b(b^x)=x$$

   This is valid for all real numbers $x$, as the domain of $f(x)=b^x$ is $(-\infty ,\infty )$.

2. Exponential of a Logarithm: Applying the exponential function after the logarithmic function.

   $$f(f^{-1}(x))=b^{{\log }_b(x)}=x$$

   This is valid only for all $x$ in the domain of $f^{-1}(x)={\log }_b(x)$, which is $x>0$.

Understanding the Domain and Range Relationship

A critical consequence of the inverse relationship between exponential and logarithmic functions is the swapping of their domains and ranges. Since the process of finding an inverse involves interchanging the input ($x$) and output ($y$) variables, the set of all possible inputs (domain) and the set of all possible outputs (range) are also interchanged.

For the Exponential Function, $f(x)=b^x$:

• Domain: The function can accept any real number as an exponent.

  $$\text{Domain: }(-\infty ,\infty )$$

• Range: The output is always a positive number. The graph has a horizontal asymptote at $y=0$ and never touches or crosses the x-axis.

  $$\text{Range: }(0,\infty )$$

For the Logarithmic Function, $f^{-1}(x)={\log }_b(x)$:

• Domain: The function can only accept positive real numbers as input. This is because the domain of the inverse is the range of the original function.

  $$\text{Domain: }(0,\infty )$$

• Range: The output (the exponent) can be any real number. This is because the range of the inverse is the domain of the original function.

  $$\text{Range: }(-\infty ,\infty )$$

This symmetrical relationship is a defining characteristic and is essential for understanding the behavior and limitations of both types of functions.

Core Concepts & Rules

• Logarithms as Inverses: The logarithmic function $g(x)={\log }_b(x)$ is, by definition, the inverse of the exponential function $f(x)=b^x$, provided the base $b$ is positive and not equal to one.

• Algebraic Inversion Process: To find the inverse of an exponential function written as $y=b^x$, you must first interchange the variables to get $x=b^y$. Then, solve for the new $y$ by converting the equation to its logarithmic form, which yields $y={\log }_b(x)$.

• Domain and Range Swap: The domain of $f(x)=b^x$, which is $(-\infty ,\infty )$, becomes the range of its inverse, $f^{-1}(x)={\log }_b(x)$. The range of $f(x)=b^x$, which is $(0,\infty )$, becomes the domain of its inverse.

• Compositional Identity: Composing an exponential function with its inverse logarithm (or vice versa) yields the original input, $x$. The two key identities are ${\log }_b(b^x)=x$ and $b^{{\log }_b(x)}=x$, each valid on the domain of the inner function.

• Graphical Symmetry: The graphs of $y=b^x$ and its inverse $y={\log }_b(x)$ are perfect reflections of one another across the line of symmetry $y=x$. Any point $(a,c)$ on the graph of the exponential function corresponds to a point $(c,a)$ on the graph of the logarithmic function.

Step-by-Step Example 1: Finding the Inverse of a Basic Exponential Function

Problem: Determine the inverse of the function $f(x)=5^x$. Then, verify the inverse relationship using composition.

Solution:

Part 1: Finding the Inverse

• Step 1: Replace $f(x)$ with $y$.

  This is standard notation for finding an inverse.

  $$y=5^x$$

• Step 2: Interchange the variables $x$ and $y$.

  This is the key algebraic step for finding any inverse function. The old output ($y$) becomes the new input ($x$), and the old input ($x$) becomes the new output ($y$). This step is based on Essential Knowledge 2.10A3.

  $$x=5^y$$

• Step 3: Solve for the new $y$.

  To isolate $y$ from the exponent, we must apply the definition of a logarithm. The equation $x=b^y$ is equivalent to $y={\log }_b(x)$. This step directly applies Essential Knowledge 2.10A1.

  $$y={\log }_5(x)$$

• Step 4: Replace $y$ with inverse function notation, $f^{-1}(x)$.

  This provides the final form of the inverse function.

  $$f^{-1}(x)={\log }_5(x)$$

Part 2: Verifying the Inverse

We use the composition property from Essential Knowledge 2.10A4 to verify our result.

• Verification 1: Check $f^{-1}(f(x))=x$.

  Substitute $f(x)=5^x$ into the inverse function.

  $$f^{-1}(f(x))={\log }_5(5^x)$$

  According to the composition property, the logarithm base 5 "undoes" the exponential base 5, leaving only the exponent.

  $${\log }_5(5^x)=x$$

  This holds true for all $x$ in the domain of $f(x)$, which is $(-\infty ,\infty )$.

• Verification 2: Check $f(f^{-1}(x))=x$.

  Substitute $f^{-1}(x)={\log }_5(x)$ into the original function.

  $$f(f^{-1}(x))=5^{({\log }_5(x))}$$

  According to the composition property, the exponential base 5 "undoes" the logarithm base 5, leaving only the input of the logarithm.

  $$5^{({\log }_5(x))}=x$$

  This holds true for all $x$ in the domain of $f^{-1}(x)$, which is $(0,\infty )$.

Since both composition tests result in $x$, we have successfully found and verified the inverse.

Step-by-Step Example 2: Graphical Analysis of an Inverse

Problem: The graph of the function $g(x)=4^x$ is provided, showing key points at $(0,1)$ and $(1,4)$. The line $y=x$ is also shown. Using this information, sketch the graph of the inverse function $g^{-1}(x)$, and state the domain and range of both $g(x)$ and $g^{-1}(x)$.

Solution:

• Step 1: Identify key features of the original function, $g(x)=4^x$.

  From the problem description and our knowledge of exponential functions (Essential Knowledge 2.10A2):

  • Key Points: $(0,1)$, $(1,4)$.

  • Asymptote: Horizontal asymptote at $y=0$.

  • Domain: $(-\infty ,\infty )$.

  • Range: $(0,\infty )$.

• Step 2: Determine key features of the inverse function, $g^{-1}(x)$.

  We use the principle that the inverse graph is a reflection over $y=x$ (Essential Knowledge 2.10A5). This means we swap the $x$ and $y$ coordinates of all features.

  • Key Points for $g^{-1}(x)$:

    • The point $(0,1)$ on $g(x)$ becomes $(1,0)$ on $g^{-1}(x)$.

    • The point $(1,4)$ on $g(x)$ becomes $(4,1)$ on $g^{-1}(x)$.

  • Asymptote for $g^{-1}(x)$:

    • The horizontal asymptote $y=0$ on $g(x)$ becomes a vertical asymptote $x=0$ on $g^{-1}(x)$.

• Step 3: Sketch the graph of $g^{-1}(x)$.

  1. Draw the vertical asymptote at $x=0$ (the y-axis).

  2. Plot the new key points: $(1,0)$ and $(4,1)$.

  3. Draw a smooth curve that passes through these points, approaches the vertical asymptote $x=0$ as $x$ approaches 0 from the right, and grows slowly as $x$ increases. The resulting curve should be a clear mirror image of the graph of $g(x)=4^x$ across the line $y=x$.

• Step 4: State the domain and range of the inverse function.

  Based on the domain/range swap rule (Essential Knowledge 2.10A2) and our graphical analysis:

  • Domain of $g^{-1}(x)$: This is the range of $g(x)$. The graph exists only for positive $x$-values.

    $$\text{Domain: }(0,\infty )$$

  • Range of $g^{-1}(x)$: This is the domain of $g(x)$. The graph extends infinitely upwards and downwards.

    $$\text{Range: }(-\infty ,\infty )$$

The inverse function is $g^{-1}(x)={\log }_4(x)$, and its graph and properties are consistent with the rules of inverses.

Using Your Calculator

The concepts in this topic are primarily analytical and graphical. A calculator is not necessary to find the inverse of $f(x)=b^x$, but it is an excellent tool for verifying your results.

Verifying an Inverse Graphically

You can use a graphing calculator to confirm that the function you found is indeed the inverse by checking for symmetry across the line $y=x$.

Example: Verify that $f^{-1}(x)={\log }_2(x)$ is the inverse of f(x) = 2^x`. 1. **Enter the Functions:** * In `Y1`, enter the original function: `Y1 = 2^X` * In `Y2`, enter the inverse function. Most calculators require the change of base formula for logarithms other than base 10 or $e. The formula is ${\log }_b(x)=\frac{\log (x)}{\log (b)}‘.So,enter:‘Y2=log(X)/log(2)$. (Some newer calculators have a $logBASE($ function in the MATH menu which can be used directly: Y2 = logBASE(2, X)).

*   In `Y3`, enter the line of reflection: `Y3 = X`

2. Set the Viewing Window:

   • For the reflection to appear accurate, the screen must have a 1:1 aspect ratio. Press ZOOM and select 5:ZSquare. This adjusts the window so that the visual distance for one unit on the x-axis is the same as one unit on the y-axis.

3. Analyze the Graph:

   • Observe the graphs of Y1 and Y2. If they are perfect mirror images of each other across the line Y3=X, your analytical work is visually confirmed. This directly checks the property from Essential Knowledge 2.10A5.

Verifying the Composition Property Numerically

You can use the calculator's table feature to check that $f(f^{-1}(x))=x$ and f^{-1}(f(x)) = x`. 1. **Enter the Compositions:** * In `Y1`, enter the first composition: `Y1 = 2^(log(X)/log(2))` * In `Y2`, enter the second composition: `Y2 = log(2^X)/log(2)` 2. **View the Table:** * Press `2nd` + `GRAPH` to access the `TABLE`. * **Analyze the `Y1` column:** You will see that for any positive `X` value, the `Y1` value is equal to $X. For $X$ values of 0 or less, you will see an ERROR. This numerically confirms that $f(f^{-1}(x))=x$ and that its domain is $(0,\infty )‘.\ast \ast \ast Analyzethe‘Y2‘column:\ast \ast Youwillseethatforall‘X‘values(positive,negative,andzero),the‘Y2‘valueisequalto‘X$. This numerically confirms that $f^{-1}(f(x))=x$ and that its domain is $(-\infty ,\infty )$.

*   This process provides strong evidence for the composition properties described in Essential Knowledge 2.10A4.

AP Exam Quick Hit

Common Question Types

• Direct Inversion: You will be given an exponential function in the form $f(x)=b^x$ and asked to find its inverse, $f^{-1}(x)$.

  • Example: "If $f(x)=10^x$, what is $f^{-1}(x)$?" (Answer: $f^{-1}(x)={\log }_{10}(x)$)

• Domain and Range Identification: You will be asked to state the domain or range of a logarithmic function, based on your knowledge that it is the inverse of an exponential function.

  • Example: "A function is defined as $g(x)={\log }_7(x)$. What is the domain of $g(x)$?" (Answer: $(0,\infty )$, because it is the range of the inverse function $f(x)=7^x$.)

• Applying Composition Properties: You will be given an expression that represents the composition of an exponential function and its inverse and asked to simplify it.

  • Example: "Simplify the expression $3^{{\log }_3(t)}$ for $t>0$." (Answer: $t$)

Common Mistakes

• Confusing Inverse with Reciprocal: A frequent error is to mistake the inverse notation $f^{-1}(x)$ for a multiplicative reciprocal. The inverse of $f(x)=b^x$ is not$b^{-x}$ or $\frac{1}{b^x}$. The inverse is the logarithmic function ${\log }_b(x)$.

• Incorrectly Applying the Logarithm Definition: When converting $x=b^y$ to logarithmic form, students might incorrectly write $y={\log }_x(b)$ or $y={\log }_b(y)$. Remember the mnemonic: "base stays the base," so the base of the exponent ($b$) becomes the base of the logarithm.

• Domain Errors on Composition: Forgetting that the identity $b^{{\log }_b(x)}=x$ is only valid for $x>0$. A question might ask for the domain over which this identity holds, and a common mistake is to state it is valid for all real numbers.

• Mixing up Domain and Range: Students often remember that the domain and range are $(-\infty ,\infty )$ and $(0,\infty )$ but forget which belongs to the exponential function and which belongs to the logarithmic function. A good way to remember is that you cannot take the logarithm of a non-positive number, so the domain of ${\log }_b(x)$ must be $(0,\infty )$.