Logarithmic Function Manipulation | undefined Unit 2 Study Guide

The Core Idea: Logarithmic Function Manipulation

Logarithmic functions possess a set of algebraic properties that are directly related to the properties of exponents. The core idea of this topic is to use these properties to manipulate and rewrite logarithmic expressions into different, but equivalent, forms. This process is fundamental for simplifying complex expressions and solving logarithmic equations.

By applying the product, quotient, and power rules, a single, complex logarithmic expression can be expanded into a sum or difference of simpler logarithms. Conversely, a series of logarithmic terms can be condensed into a single logarithm. Furthermore, the change of base formula provides a crucial tool for evaluating logarithms of any base using a standard calculator and for re-expressing logarithms in a more convenient base for analysis. Mastering these manipulations is essential for working proficiently with logarithmic functions.

Key Logarithmic Properties

The following properties are used to rewrite logarithmic expressions in equivalent forms. For all rules, the bases $b$ and $c$ must be positive and not equal to 1 ( $b > 0, b \neq = 1, c > 0, c \neq = 1$ ), and the arguments $x$ and $y$ must be positive ( $x > 0, y > 0$ ).

1. The Product Rule

This rule relates the logarithm of a product to the sum of the logarithms of its factors.

$lo g_{b} (x y) = lo g_{b} (x) + lo g_{b} (y)$

2. The Quotient Rule

This rule relates the logarithm of a quotient to the difference of the logarithms of the numerator and denominator.

$lo g_{b} (\frac{x}{y}) = lo g_{b} (x) - lo g_{b} (y)$

3. The Power Rule

This rule allows an exponent on the argument of a logarithm to be rewritten as a coefficient of the logarithm.

$lo g_{b} (x^{p}) = p \cdot lo g_{b} (x)$

4. The Change of Base Formula

This formula allows a logarithm with any base $b$ to be expressed as a ratio of logarithms with a new base $c$ . This is particularly useful for calculations, as calculators typically only have keys for base 10 ( $l o g$ ) and base $e$ ( $l n$ ).

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

Understanding Domain Restrictions

A critical aspect of applying the logarithmic properties is understanding the conditions under which they are valid. The Essential Knowledge statements consistently specify that the arguments of the logarithms, $x$ and $y$ , must be positive ( $x > 0$ , $y > 0$ ). This is a direct consequence of the definition of a logarithmic function, $y = l o g_{b} (x)$ , which is only defined for positive values of $x$ .

When we manipulate an expression like $l o g_{b} (x y)$ , the original expression is defined only when the product $x y$ is positive. However, when we expand it to $l o g_{b} (x) + l o g_{b} (y)$ , the expanded form requires both $x$ and $y$ to be individually positive. The rules provided in the AP Precalculus curriculum are stated under these stricter conditions ( $x > 0$ and $y > 0$ ). Therefore, when applying these properties, you must operate under the assumption that all variables in the arguments of the logarithms represent positive numbers. This ensures that every term in both the original and manipulated expressions is well-defined.

Core Concepts & Rules

Expanding Products: The logarithm of a product of two positive numbers is equivalent to the sum of their individual logarithms.
Expanding Quotients: The logarithm of a quotient of two positive numbers is equivalent to the logarithm of the numerator minus the logarithm of the denominator.
Manipulating Exponents: An exponent on the argument of a logarithm can be moved to the front of the expression as a multiplier.
Condensing Sums: A sum of two logarithms with the same base can be combined into a single logarithm of their arguments' product.
Condensing Differences: A difference of two logarithms with the same base can be combined into a single logarithm of their arguments' quotient.
Changing Bases: Any logarithm can be evaluated or rewritten by dividing the logarithm of the argument by the logarithm of the base, using any new, valid base.
Domain is Key: All logarithmic properties are defined under the condition that the arguments of all logarithms involved are positive numbers.

Step-by-Step Example 1: Expanding and Condensing Logarithmic Expressions

This example demonstrates how to use the product, quotient, and power rules to rewrite logarithmic expressions.

Part A: Expanding an Expression

Write the expression $lo g_{5} (\frac{x ^{3} y}{z ^{4}})$ as a sum or difference of logarithms, where $x$ , $y$ , and $z$ are positive numbers.

Step 1: Apply the Quotient Rule

The main operation is division, so we start by separating the numerator and the denominator.

$lo g_{5} (\frac{x ^{3} y}{z ^{4}}) = lo g_{5} (x^{3} y) - lo g_{5} (z^{4})$

Step 2: Apply the Product Rule

The first term contains a product, $x^{3}$ times $y$ . We can expand this term.

$lo g_{5} (x^{3} y) - lo g_{5} (z^{4}) = lo g_{5} (x^{3}) + lo g_{5} (y) - lo g_{5} (z^{4})$

Step 3: Apply the Power Rule

Rewrite the square root as an exponent ( $y = y^{1/2}$ ). Now, apply the power rule to all three terms to move the exponents to the front as coefficients.

$lo g_{5} (x^{3}) + lo g_{5} (y^{1/2}) - lo g_{5} (z^{4}) = 3 lo g_{5} (x) + \frac{1}{2} lo g_{5} (y) - 4 lo g_{5} (z)$

This is the fully expanded form.

Part B: Condensing an Expression

Write the expression $2 lo g_{b} (x) - \frac{1}{3} lo g_{b} (y) + 5 lo g_{b} (z)$ as a single logarithm.

Step 1: Apply the Power Rule in Reverse

Move the coefficients back to become exponents on the arguments of their respective logarithms.

$lo g_{b} (x^{2}) - lo g_{b} (y^{1/3}) + lo g_{b} (z^{5})$

Step 2: Apply the Quotient and Product Rules

Work from left to right. First, combine the difference into a quotient.

$lo g_{b} (\frac{x ^{2}}{y ^{1/3}}) + lo g_{b} (z^{5})$

Next, combine the sum into a product. The term $z^{5}$ will be in the numerator.

$lo g_{b} (\frac{x ^{2} \cdot z ^{5}}{y ^{1/3}})$

This is the fully condensed form, which can also be written with a cube root in the denominator: $lo g_{b} (\frac{x ^{2} z ^{5}}{3 y})$ .

Step-by-Step Example 2: Applying the Change of Base Formula

Suppose you are given that $lo g_{3} (10) \approx 2.096$ and $lo g_{3} (2) \approx 0.631$ . Use the change of base formula and the given values to approximate the value of $lo g_{5} (10)$ .

Step 1: Choose a New Base

The problem provides values for logarithms in base 3. This suggests that we should use $c = 3$ as the new base in the change of base formula.

Step 2: Apply the Change of Base Formula

The expression we want to evaluate is $lo g_{5} (10)$ . Using the formula $lo g_{b} (x) = \frac{l o g _{c} ( x )}{l o g _{c} ( b )}$ , we set $b = 5$ , $x = 10$ , and $c = 3$ .

$lo g_{5} (10) = \frac{lo g _{3} ( 10 )}{lo g _{3} ( 5 )}$

Step 3: Manipulate the Expression to Use Given Values

We are given $lo g_{3} (10)$ , but we need $lo g_{3} (5)$ . We can create $lo g_{3} (5)$ by recognizing that $5 = 10/2$ . We can use the quotient rule.

$lo g_{3} (5) = lo g_{3} (\frac{10}{2}) = lo g_{3} (10) - lo g_{3} (2)$

Step 4: Substitute the Given Approximations

Now, substitute the given numerical values to find an approximation for $lo g_{3} (5)$ .

$lo g_{3} (5) \approx 2.096 - 0.631 = 1.465$

Step 5: Calculate the Final Result

Substitute the values for $lo g_{3} (10)$ and our calculated value for $lo g_{3} (5)$ back into the change of base formula from Step 2.

$lo g_{5} (10) = \frac{lo g _{3} ( 10 )}{lo g _{3} ( 5 )} \approx \frac{2.096}{1.465} \approx 1.431$

Therefore, $lo g_{5} (10) \approx 1.431$ .

Using Your Calculator

The most direct application of a calculator for this topic is evaluating logarithms with bases other than 10 or $e$ using the Change of Base Formula.

Problem: Calculate the value of $lo g_{7} (200)$ to four decimal places.

Method: Your calculator has a LOG button (for base 10) and an LN$ button (for base $e$ ). You can use either one for the change of base formula. We will demonstrate with LN$ (base $e$ ).

Step 1: Identify $b$ and $x$

In the expression $lo g_{7} (200)$ , the base is $b = 7$ and the argument is $x = 200$ .

Step 2: Set up the Change of Base Formula

Using the natural logarithm ( $l n$ , base $e$ ) as our new base $c$ , the formula is:

$lo g_{7} (200) = \frac{ln ( 200 )}{ln ( 7 )}$

Step 3: Enter into the Calculator

On a TI-84 style calculator, you would type the following sequence:

Press the LN button.
Type $200$ .
Press the $)$ button to close the parenthesis.
Press the $\div$ (division) button.
Press the LN button.
Type $7$ .
Press the $)$ button.
Press ENTER.

The calculator screen would look like: $l n (200) / l n (7)$

Step 4: Record the Result

The calculator will display approximately $2.722706...$ . Rounding to four decimal places, the answer is $2.7227$ .

Note: Using the common logarithm (`LOG$, base 10) would yield the same result: $l o g (200) / l o g (7) \approx 2.7227$ .

AP Exam Quick Hit

Common Question Types

Expanding/Condensing Expressions: You will be given a logarithmic expression and asked to choose an equivalent form from a list of options.
- Example: "Which of the following is equivalent to $ln (\frac{5 x ^{2}}{y})$ for $x > 0$ and $y > 0$ ?"
- (A) $2 ln (5 x) - ln (y)$
- (B) $ln (5) + 2 ln (x) - ln (y)$
- (C) $ln (5) + ln (x^{2}) + ln (y)$
- (D) $ln (5) \cdot 2 ln (x) \div ln (y)$
Evaluating Logarithms Using Given Values: You will be provided with the values of a few simple logarithms and asked to find the value of a more complex logarithm by breaking it down into its factors.
- Example: "If $lo g_{b} (2) \approx 0.356$ and $lo g_{b} (3) \approx 0.565$ , what is the approximate value of $lo g_{b} (24)$ ?" (Hint: $24 = 8 \cdot 3 = 2^{3} \cdot 3$ )
Applying the Change of Base Formula: You may be asked to evaluate a logarithm with an uncommon base, requiring a calculator, or to identify the correct change of base setup.
- Example: "Which of the following expressions can be used to calculate $lo g_{6} (15)$ ?"
- (A) $\frac{l n ( 6 )}{l n ( 15 )}$
- (B) $ln (15) - ln (6)$
- (C) $\frac{l o g ( 15 )}{l o g ( 6 )}$
- (D) $lo g (15 - 6)$

Common Mistakes

Confusing the Log of a Sum with the Sum of Logs: A very common error is to assume $lo g_{b} (x + y)$ is the same as $lo g_{b} (x) + lo g_{b} (y)$ . The product rule applies to the log of a product, $lo g_{b} (x y)$ , not a sum.
Incorrectly Applying the Power Rule: Students often mistake $(lo g_{b} (x))^{p}$ for $lo g_{b} (x^{p})$ . The power rule $p \cdot lo g_{b} (x)$ only applies when the exponent $p$ is on the argument $x$ , not on the entire logarithmic expression.
Errors in the Quotient Rule: Reversing the order of subtraction is a frequent mistake. Remember that $lo g_{b} (x) - lo g_{b} (y)$ corresponds to $lo g_{b} (x / y)$ , with the positive term's argument in the numerator and the negative term's argument in the denominator.
Flipping the Change of Base Formula: A common mistake is to write the formula as $\frac{l o g _{c} ( b )}{l o g _{c} ( x )}$ instead of the correct $\frac{l o g _{c} ( x )}{l o g _{c} ( b )}$ . Remember: the original argument ( $x$ ) goes in the numerator, and the original base ( $b$ ) goes in the denominator.

Logarithmic Function Manipulation - AP PreCalculus Study Guide