Logarithmic Functions - AP PreCalculus Study Guide

The Core Idea: Logarithmic Functions

Logarithmic functions are introduced as the inverse of exponential functions. While an exponential function takes a base and an exponent as input to find a resulting value (e.g., $2^3=8$), a logarithmic function does the reverse. It takes a base and a resulting value as input to find the original exponent (e.g., ${\log }_2(8)=3$). The fundamental purpose of a logarithmic function is to answer the question: "To what power must we raise the base to obtain a specific number?"

The function $f(x)={\log }_b(x)$ is formally defined by its relationship to the exponential function $x=b^y$. The output of the logarithm, $y$, is precisely the exponent needed. This inverse relationship dictates the core properties of logarithmic functions, including their domain, which is restricted to positive numbers, and the necessary conditions for their base.

Key Formulas/Rules

The single most important relationship for logarithmic functions is the equivalence between logarithmic and exponential forms. This equivalence is the basis for evaluating and solving all logarithmic expressions and equations.

Logarithmic-Exponential Equivalence

The statement $y={\log }_b(x)$ is equivalent to the statement $x=b^y$.

• $y={\log }_b(x)$ is the logarithmic form.

• $x=b^y$ is the exponential form.

Components of the Logarithmic Function:

• $b$ is the base: The base must be a positive number other than 1 ($b>0$ and $b\ne 1$).

• $x$ is the argument: The argument is the input to the function. The argument must be a positive number ($x>0$).

• $y$ is the output (the logarithm): The output is the exponent to which the base $b$ must be raised to produce the argument $x$.

Understanding the Domain and Base

The definitions and constraints of logarithmic functions are a direct consequence of their nature as inverses of exponential functions. Understanding these constraints is critical for working with logarithms correctly.

The Domain of a Logarithmic Function

The domain of the parent logarithmic function $f(x)={\log }_b(x)$ is the set of all positive real numbers, or $x>0$. In interval notation, this is $(0,\infty )$.

• Why? This restriction comes from the exponential relationship $x=b^y$. Since the base $b$ must be a positive number, raising a positive number to any real exponent $y$ will always result in a positive number $x$. It is impossible for $b^y$ to be zero or negative. Therefore, the input `x$ of the corresponding logarithmic function, ${\log }_b(x)$, must be positive.

The Range of a Logarithmic Function

The range of the parent logarithmic function $f(x)={\log }_b(x)$ is the set of all real numbers. In interval notation, this is $(-\infty ,\infty )$.

• Why? The output of the logarithm, $y$, is an exponent. This exponent can be any real number—positive (e.g., $2^3=8$), negative (e.g., $2^{-3}=1/8$), or zero (e.g., $2^0=1$). Since the exponent $y$ in $x=b^y$ can be any real number, the range of its inverse, $f(x)={\log }_b(x)$, is all real numbers.

Restrictions on the Base $b$

The base $b$ of a logarithmic function must be a positive number and cannot be equal to 1 ($b>0,b\ne 1$).

• Why must $b$ be positive? This is a convention inherited from the definition of exponential functions. It ensures the function is well-defined and avoids issues with even roots of negative numbers (e.g., $\sqrt{-4}$).

• Why can't $b$ be 1? If the base were 1, the exponential form would be $x=1^y$. Since 1 raised to any power is always 1, this would mean $x$ is always 1. The function would be a vertical line $x=1$, not a one-to-one function that can have a proper inverse. The corresponding "logarithm" ${\log }_1(x)$ would only be defined for $x=1$ and could be any real number, making it not a function.

Core Concepts & Rules

• Inverse Relationship: A logarithmic function is the inverse of an exponential function. The function $f(x)={\log }_b(x)$ "undoes" the function $g(x)=b^x$.

• Converting Forms: To evaluate or solve a logarithmic expression, convert it to its equivalent exponential form. The expression $y={\log }_b(x)$ is identical in meaning to $x=b^y$.

• The Output is an Exponent: The value of ${\log }_b(x)$ is the exponent you must place on the base $b$ to get the argument $x$.

• Domain Restriction: The input (argument) of any logarithmic function must be strictly positive. For $f(x)={\log }_b(g(x))$, the domain is determined by the inequality $g(x)>0$.

• Range: The output of a parent logarithmic function can be any real number.

• Base Requirements: The base of a logarithm, $b$, must be positive and not equal to one ($b>0,b\ne 1$).

Step-by-Step Example 1: Evaluating a Logarithm

Problem: Determine the value of ${\log }_4(64)$.

Step 1: Set the expression equal to a variable.

Let the unknown value be $y$.

$$y={\log }_4(64)$$

Step 2: Convert the logarithmic equation to its equivalent exponential form.

Using the relationship $y={\log }_b(x)⟺x=b^y$, we identify:

• $b=4$ (the base)

• $x=64$ (the argument)

• $y=y$ (the unknown exponent)

The exponential form is:

$$64=4^y$$

Step 3: Solve for the variable.

The question now is: "To what power must 4 be raised to get 64?" We can solve this by finding a common base or by inspection.

• $4^1=4$

• $4^2=16$

• $4^3=64$

We see that $y=3$.

Step 4: State the final answer.

Therefore, ${\log }_4(64)=3$.

Step-by-Step Example 2: Solving for an Input

Problem: A function is defined as $f(x)={\log }_5(x)$. Determine the input value $x$ for which the output is $f(x)=-2$.

Step 1: Set up the equation.

We are given that $f(x)=-2$. Substitute this into the function definition:

$$-2={\log }_5(x)$$

Step 2: Convert the logarithmic equation to its equivalent exponential form.

Identify the components:

• $b=5$ (the base)

• $x=x$ (the unknown argument)

• $y=-2$ (the output/exponent)

The exponential form is:

$$x=5^{-2}$$

Step 3: Evaluate the exponential expression to find $x$

Recall the rule of negative exponents, $a^{-n}=\frac{1}{a^n}$.

$$x=\frac{1}{5^2}$$

$$x=\frac{1}{25}$$

Step 4: Verify the answer.

The input `x = \frac{1}{25}must be in the domain of $f(x) = \log_5(x). The domain requires $x>0$. Since $\frac{1}{25}>0$, the solution is valid.

The input value is $x=\frac{1}{25}$.

Using Your Calculator

While the conceptual understanding of logarithms relies on the exponential equivalence, a calculator is a practical tool for determining outputs (evaluating) for bases that are not simple integers.

Problem: Determine the output of the function $f(x)={\log }_3(30)$ to three decimal places.

Method 1: Using the $logBASE($ Function (TI-84 and similar models)

1. Press the $[alpha]$ key, then the $[window]$ key to bring up the function shortcut menu.

2. Select option 5: $logBASE($.

3. The calculator will display a template: ${\log }_{\square }(\square )$.

4. Enter the base in the first box: $3$.

5. Enter the argument in the second box: $30$.

6. Your screen should show ${\log }_3(30)$.

7. Press `[enter]. The result will be approximately $3.0959....

8. Round to three decimal places: $3.096$.

Method 2: Using the Change of Base Formula (Any Scientific/Graphing Calculator)

The change of base formula states that ${\log }_b(x)=\frac{\log (x)}{\log (b)}=\frac{\ln (x)}{\ln (b)}$. Most calculators have dedicated buttons for $log$ (base 10) and $ln$ (base $e$).

1. Press the $[log]$ button.

2. Enter the argument: $30$.

3. Close the parenthesis: $)$.

4. Press the division $[/]$ button.

5. Press the $[log]$ button again.

6. Enter the base: $3$.

7. Close the parenthesis: $)$.

8. Your screen should show $log(30)/log(3)$.

9. Press `[enter]. The result will be approximately $3.0959....

10. Round to three decimal places: $3.096$.

Both methods confirm that $f(30)\approx 3.096$.

AP Exam Quick Hit

Common Question Types

• Direct Evaluation: You will be asked to find the value of a logarithmic expression without a calculator.

  • Example: "What is the value of ${\log }_2\left(\frac{1}{16}\right)$?" (Answer: -4, because $2^{-4}=\frac{1}{16}$)

• Solving for an Input or Base: You will be given a logarithmic equation and asked to solve for the argument ($x$) or the base ($b$).

  • Example: "If ${\log }_b(125)=3$, what is the value of $b$?" (Answer: 5, because $b^3=125$)

• Determining Domain: You will be given a logarithmic function with a transformation and asked to identify its domain.

  • Example: "What is the domain of the function $g(x)={\log }_7(x-4)$?" (Answer: $x>4$, because the argument $x-4$ must be greater than 0)

Common Mistakes

• Mixing up Components: Incorrectly converting between logarithmic and exponential form. A common error is swapping the base and the argument, for example, converting ${\log }_2(8)=y$ to $y^2=8$ instead of the correct $2^y=8$.

• Forgetting the Domain Restriction: Attempting to evaluate a logarithm with a non-positive argument, such as ${\log }_3(-9)$, or failing to correctly state the domain of a transformed logarithmic function. The argument of a logarithm can never be zero or negative.

• Ignoring Base Restrictions: Forgetting that the base of a logarithm must be positive and not equal to 1. This can lead to incorrect answers when solving for an unknown base.

• Arithmetic with Negative Exponents: Making errors when evaluating the exponential form for a negative output. For example, incorrectly calculating $4^{-2}$ as $-16$ or $-8$ instead of the correct $\frac{1}{16}$.