Logarithmic Expressions | undefined Unit 2 Study Guide

The Core Idea: Logarithmic Expressions

The fundamental concept of a logarithm is to provide a name for an exponent. The expression $lo g_{b} (x)$ answers the question: "To what exponent must the base $b$ be raised to produce the value $x$ ?" This establishes a direct relationship between logarithmic and exponential forms. Because logarithms are fundamentally exponents, they follow a set of properties that are derived directly from the properties of exponents.

These properties—the Product, Quotient, and Power Properties—are not arbitrary rules but are the logarithmic equivalent of the rules for multiplying, dividing, and raising powers of exponential expressions with the same base. Mastering these properties allows us to rewrite and simplify complex logarithmic expressions, either by expanding a single logarithm into multiple terms or by condensing multiple terms into a single logarithm. This skill is crucial for solving logarithmic equations and analyzing logarithmic functions. Furthermore, the change of base formula provides a practical tool for evaluating any logarithm using standard calculator functions.

Key Formulas & Properties

The following properties are used to rewrite and simplify logarithmic expressions. They are all derived from the fundamental properties of exponents.

The Definition of a Logarithm

The logarithm $lo g_{b} (x)$ is defined as the exponent, $y$ , to which the base $b$ must be raised to produce $x$ . This can be written as an equivalence:

$lo g_{b} (x) = y ⟺ b^{y} = x$

Properties of Logarithms

These properties apply for any valid base $b$ and positive arguments $x$ and $y$ .

Product Property: The logarithm of a product is the sum of the logarithms of its factors. This is based on the exponent rule $b^{m} \cdot b^{n} = b^{m + n}$ .
$lo g_{b} (x y) = lo g_{b} (x) + lo g_{b} (y)$
Quotient Property: The logarithm of a quotient is the difference of the logarithms of the numerator and the denominator. This is based on the exponent rule $\frac{b ^{m}}{b ^{n}} = b^{m - n}$ .
$lo g_{b} (\frac{x}{y}) = lo g_{b} (x) - lo g_{b} (y)$
Power Property: The logarithm of a number raised to a power is the product of the power and the logarithm of the number. This is based on the exponent rule $(b^{m})^{p} = b^{m p}$ .
$lo g_{b} (x^{p}) = p \cdot lo g_{b} (x)$

Change of Base Formula

This formula allows for the evaluation of a logarithm with any base by converting it into a quotient of logarithms with a different, more convenient base (such as base 10 or base $e$ , which are available on calculators).

$lo g_{b} (x) = \frac{lo g _{c} ( x )}{lo g _{c} ( b )}$

Here, $c$ can be any valid new base.

Understanding the Exponent Connection

The properties of logarithms are not new rules to memorize but are direct consequences of the properties of exponents. Understanding this connection is essential for truly mastering logarithmic expressions. The core idea is that since $lo g_{b} (x)$ is an exponent, manipulating logarithms is the same as manipulating exponents.

Let's formally establish this connection for each property:

1. The Product Property: $lo g_{b} (x y) = lo g_{b} (x) + lo g_{b} (y)$

Let $m = lo g_{b} (x)$ and $n = lo g_{b} (y)$ .
By the definition of a logarithm, these can be rewritten in exponential form as $b^{m} = x$ and $b^{n} = y$ .
Now consider the product $x y$ . We can substitute our exponential forms: $x y = b^{m} \cdot b^{n}$ .
Using the property of exponents for multiplication, we get $x y = b^{m + n}$ .
Now, let's convert this back to logarithmic form. The base is $b$ , the result is $x y$ , and the exponent is $m + n$ . So, $lo g_{b} (x y) = m + n$ .
Finally, substitute the original definitions of $m$ and $n$ back in: $lo g_{b} (x y) = lo g_{b} (x) + lo g_{b} (y)$ .

2. The Quotient Property: $lo g_{b} (x / y) = lo g_{b} (x) - lo g_{b} (y)$

Using the same setup, $m = lo g_{b} (x)$ (so $b^{m} = x$ ) and $n = lo g_{b} (y)$ (so $b^{n} = y$ ).
Consider the quotient $x / y$ . Substitute the exponential forms: $x / y = b^{m} / b^{n}$ .
Using the property of exponents for division, we get $x / y = b^{m - n}$ .
Converting this back to logarithmic form gives $lo g_{b} (x / y) = m - n$ .
Substituting back for $m$ and $n$ gives: $lo g_{b} (x / y) = lo g_{b} (x) - lo g_{b} (y)$ .

3. The Power Property: $lo g_{b} (x^{p}) = p \cdot lo g_{b} (x)$

Let $m = lo g_{b} (x)$ , which means $b^{m} = x$ .
Consider the expression $x^{p}$ . Substitute the exponential form of $x$ : $x^{p} = (b^{m})^{p}$ .
Using the property of exponents for powers, we get $x^{p} = b^{m p}$ .
Converting this final exponential statement back to logarithmic form gives $lo g_{b} (x^{p}) = m p$ .
Substituting the original definition of $m$ gives: $lo g_{b} (x^{p}) = p \cdot lo g_{b} (x)$ .

Core Concepts & Rules

Logarithms are Exponents: The expression $lo g_{b} (x)$ represents the power you must raise base $b$ to in order to get $x$ .
Properties Mirror Exponent Rules: The rules for manipulating logarithms are a direct result of the rules for exponents. Adding logs corresponds to multiplying arguments; subtracting logs corresponds to dividing arguments; and multiplying a log by a constant corresponds to raising the argument to that power.
Expanding Expressions: A single logarithm containing products, quotients, or powers in its argument can be broken down into a sum, difference, or multiple of simpler logarithms.
Condensing Expressions: A series of logarithmic terms (with the same base) connected by addition and subtraction can be combined into a single logarithm by applying the properties in reverse.
Universal Evaluation: The change of base formula is a powerful tool that allows you to calculate the value of a logarithm of any base using a calculator, which typically only has keys for base 10 ( $lo g$ ) and base $e$ ( $ln$ ).

Step-by-Step Example 1: Expanding a Logarithmic Expression

Problem: Fully expand the expression $lo g_{5} (\frac{25 x}{y ^{3}})$ .

Step 1: Apply the Quotient Property

The main operation in the argument is division. We can split the logarithm of the quotient into a difference of two logarithms.

$lo g_{5} (\frac{25 x}{y ^{3}}) = lo g_{5} (25 x) - lo g_{5} (y^{3})$

Step 2: Apply the Product Property

The first term, $lo g_{5} (25 x)$ , contains a product. We can split this into a sum of two logarithms.

$lo g_{5} (25) + lo g_{5} (x) - lo g_{5} (y^{3})$

Step 3: Rewrite Radicals as Exponents and Apply the Power Property

Rewrite the square root as a fractional exponent: $x = x^{1/2}$ . Now, apply the power property to any term with an exponent in its argument. The exponent moves to the front as a coefficient.

$lo g_{5} (25) + \frac{1}{2} lo g_{5} (x) - 3 lo g_{5} (y)$

Step 4: Evaluate and Simplify

Check if any of the logarithmic terms can be evaluated. We know that $5^{2} = 25$ , so by definition, $lo g_{5} (25) = 2$ .

$2 + \frac{1}{2} lo g_{5} (x) - 3 lo g_{5} (y)$

This is the fully expanded form of the original expression.

Step-by-Step Example 2: Condensing and Evaluating an Expression

Problem: Rewrite the expression $2 lo g_{3} (6) - \frac{1}{2} lo g_{3} (16)$ as a single logarithm and find its value.

Step 1: Apply the Power Property in Reverse

The coefficients in front of the logarithmic terms can be moved back into the argument as exponents.

$2 lo g_{3} (6) \to lo g_{3} (6^{2}) = lo g_{3} (36)$

$\frac{1}{2} lo g_{3} (16) \to lo g_{3} (1 6^{1/2}) = lo g_{3} (16) = lo g_{3} (4)$

The expression is now:

$lo g_{3} (36) - lo g_{3} (4)$

Step 2: Apply the Quotient Property in Reverse

A difference of two logarithms with the same base can be condensed into a single logarithm of a quotient.

$lo g_{3} (36) - lo g_{3} (4) = lo g_{3} (\frac{36}{4})$

Step 3: Simplify the Argument

Perform the division inside the argument of the logarithm.

$lo g_{3} (9)$

Step 4: Evaluate the Final Logarithm

Ask the question: "To what power must 3 be raised to get 9?" Since $3^{2} = 9$ , the answer is 2.

$lo g_{3} (9) = 2$

The value of the original expression is 2.

Using Your Calculator

The primary use of a calculator for this topic is to evaluate logarithms that have bases other than 10 or $e$ , using the Change of Base Formula. Most scientific and graphing calculators have dedicated keys for $lo g$ (common logarithm, base 10) and $ln$ (natural logarithm, base $e$ ).

Problem: Calculate the value of $lo g_{6} (80)$ rounded to three decimal places.

Method:

Identify the base ( $b = 6$ ) and the argument ( $x = 80$ ).
Choose a new base to convert to, typically base 10 ( $lo g$ ) or base $e$ ( $ln$ ). Both will yield the same result. Let's use base 10.
Apply the Change of Base Formula: $lo g_{b} (x) = \frac{l o g _{c} ( x )}{l o g _{c} ( b )}$ .
$lo g_{6} (80) = \frac{lo g _{10} ( 80 )}{lo g _{10} ( 6 )} = \frac{lo g ( 80 )}{lo g ( 6 )}$

TI-84 Style Keystrokes:

Press the $[l o g]$ key.
Enter the argument of the numerator: $80$ .
Close the parenthesis: $)$ .
Press the division key: $[/]$ .
Press the $[l o g]$ key again.
Enter the argument of the denominator (the original base): $6$ .
Close the parenthesis: $)$ .
Press [ENTER].

The calculator will display approximately $2.4453...$ . Rounded to three decimal places, the answer is 2.445.

Note: Some newer calculators have a $l o g B A SE ($ function (often found in the MATH menu) that allows you to enter the base and argument directly. While convenient, you must know how to use the Change of Base formula as it is a required piece of knowledge.

AP Exam Quick Hit

Common Question Types

Expanding/Condensing Expressions: You will be given a logarithmic expression and asked to find an equivalent expression among the multiple-choice options. This is a direct test of the product, quotient, and power properties.
- Example: "Which of the following is equivalent to $ln (\frac{x ^{3}}{e ^{2} y})$ ?"
- Answer: $ln (x^{3}) - ln (e^{2} y) = 3 ln (x) - (ln (e^{2}) + ln (y)) = 3 ln (x) - (2 + ln (y)) = 3 ln (x) - 2 - ln (y)$
Evaluating Numerical Expressions: You will be asked to find the numerical value of a logarithmic expression, which may require condensing first or using the change of base formula.
- Example: "What is the value of $lo g_{4} (2) + lo g_{4} (32)$ ?"
- Answer: $lo g_{4} (2 \cdot 32) = lo g_{4} (64) = 3$ , since $4^{3} = 64$ .
Solving for Variables in Terms of Other Variables: You are given the values of simple logarithms in terms of variables and asked to find the value of a more complex logarithm.
- Example: "If $A = lo g (2)$ and $B = lo g (3)$ , what is $lo g (180)$ in terms of A and B?"
- Answer: $lo g (180) = lo g (18 \cdot 10) = lo g (18) + lo g (10) = lo g (2 \cdot 3^{2}) + 1 = lo g (2) + lo g (3^{2}) + 1 = lo g (2) + 2 lo g (3) + 1 = A + 2 B + 1$

Common Mistakes

Confusing Log of a Sum with Sum of Logs: A very common error is to assume $lo g_{b} (x + y)$ can be simplified to $lo g_{b} (x) + lo g_{b} (y)$ . This is incorrect. The sum of logs property only applies to the log of a product, $lo g_{b} (x y)$ .
Incorrectly Applying the Power Rule: When condensing an expression like $3 lo g (x + 2)$ , students might incorrectly apply the 3 to just the $x$ , resulting in $lo g (x^{3} + 2)$ . The correct application is to the entire argument: $lo g ((x + 2)^{3})$ .
Flipping the Change of Base Formula: When calculating $lo g_{b} (x)$ , a frequent mistake is to compute $\frac{l o g ( b )}{l o g ( x )}$ instead of the correct $\frac{l o g ( x )}{l o g ( b )}$ . A helpful mnemonic is "the base goes in the basement" (the denominator).
Ignoring the Base: Attempting to condense expressions with different bases, such as $lo g_{2} (x) + lo g_{4} (y)$ , without first changing them to a common base. The properties of logarithms only apply when the bases are identical.
Errors with Coefficients and Subtraction: In an expression like $lo g (x) - 2 lo g (y)$ , a common mistake is to condense this to $lo g (\frac{x}{y ^{2}})$ correctly, but in more complex forms, students might misapply the subtraction, for instance, with $lo g (x) - (lo g (y) + lo g (z))$ , forgetting to distribute the negative to get $lo g (x) - lo g (y) - lo g (z)$ .

Logarithmic Expressions - AP PreCalculus Study Guide