Applying the Laws of Logarithms, Including Change of Base

Introduction

Key Concepts

The Definition of a Logarithm

A logarithm is the inverse operation of exponentiation. It answers the question: "To what power must a specific base be raised to produce a given number?" Mathematically, for any positive real numbers \( a \), \( b \), and \( c \), where \( a \neq 1 \), the logarithm satisfies the equation:

$$\log_a b = c \iff a^c = b$$

The Fundamental Laws of Logarithms

• Product Law: The logarithm of a product is equal to the sum of the logarithms of the factors:

  $$\log_a (BC) = \log_a B + \log_a C$$

• Quotient Law: The logarithm of a quotient is equal to the difference of the logarithms:

  $$\log_a \left(\frac{B}{C}\right) = \log_a B - \log_a C$$

• Power Law: The logarithm of a number raised to a power is equal to the exponent times the logarithm of the number:

  $$\log_a (B^k) = k \log_a B$$

Change of Base Formula

Sometimes, it's necessary to compute logarithms with bases other than 10 or \( e \). The change of base formula allows logarithms to be converted to any other base, typically to simplify calculations:

$$\log_b a = \frac{\log_c a}{\log_c b}$$

Typically, base 10 (\( \log \)) or base \( e \) (\( \ln \)) are used for convenience.

Solving Logarithmic Equations

Using the laws of logarithms, we can solve various logarithmic equations. For example, to solve:

$$\log_2 (x) + \log_2 (x - 1) = 3$$

Apply the product law:

$$\log_2 (x(x - 1)) = 3$$

Therefore:

$$x(x - 1) = 2^3 = 8$$

Simplifying gives the quadratic equation:

$$x^2 - x - 8 = 0$$

The solutions are:

$$x = \frac{1 \pm \sqrt{1 + 32}}{2} = \frac{1 \pm \sqrt{33}}{2}$$

Only the positive solution is valid since logarithms of positive numbers are defined.

Logarithmic Identities

Several important identities facilitate the manipulation of logarithmic expressions:

• Identity 1: \( \log_a 1 = 0 \) because \( a^0 = 1 \).
• Identity 2: \( \log_a a = 1 \) because \( a^1 = a \).
• Identity 3: \( \log_a (a^k) = k \).

These identities are often used in simplifying logarithmic expressions and solving equations.

Exponential and Logarithmic Relationships

Understanding the relationship between exponential and logarithmic functions is crucial. They are inverse functions, meaning that:

$$a^{\log_a b} = b$$

and

$$\log_a (a^b) = b$$

This inverse relationship allows for the conversion between exponential and logarithmic forms, which is often necessary in solving equations.

Examples and Applications

Consider the following example to illustrate the application of logarithmic laws:

Example: Solve for \( x \) in the equation \( 3 \log_2 x - \log_2 4 = 2 \).

Solution:

First, apply the power law to the first term:

$$3 \log_2 x = \log_2 x^3$$

Then, substitute back into the equation:

$$\log_2 x^3 - \log_2 4 = 2$$

Apply the quotient law:

$$\log_2 \left(\frac{x^3}{4}\right) = 2$$

Convert to exponential form:

$$\frac{x^3}{4} = 2^2 = 4$$

Solve for \( x \):

$$x^3 = 16 \implies x = \sqrt[3]{16}$$

Advanced Concepts

Deriving the Change of Base Formula

The change of base formula can be derived using the definition of logarithms and the properties of exponents. Starting with:

$$b^c = a$$

Take the logarithm of both sides to a common base \( d \):

$$\log_d b^c = \log_d a$$

Applying the power law:

$$c \log_d b = \log_d a$$

Thus:

$$c = \frac{\log_d a}{\log_d b}$$

Since \( c = \log_b a \), we have:

$$\log_b a = \frac{\log_d a}{\log_d b}$$

This establishes the change of base formula.

Solving Complex Logarithmic Equations

For equations that involve multiple logarithmic terms and require the application of several laws simultaneously, the process becomes more intricate. Consider the equation:

$$\log_3 (x) + 2\log_3 (x + 4) = \log_3 (3x + 12)$$

Solution:

First, apply the power law to the second term:

$$\log_3 (x) + \log_3 (x + 4)^2 = \log_3 (3x + 12)$$

Next, apply the product law:

$$\log_3 [x(x + 4)^2] = \log_3 (3x + 12)$$

Since the logarithms are equal, their arguments must be equal:

$$x(x + 4)^2 = 3x + 12$$

Expanding the left side:

$$x(x^2 + 8x + 16) = x^3 + 8x^2 + 16x$$

Set the equation to zero:

$$x^3 + 8x^2 + 16x - 3x - 12 = 0$$

Combine like terms:

$$x^3 + 8x^2 + 13x - 12 = 0$$

Solving this cubic equation may require methods such as factoring by grouping or using the Rational Root Theorem. Suppose \( x = 1 \) is a root:

$$1^3 + 8(1)^2 + 13(1) - 12 = 1 + 8 + 13 - 12 = 10 \neq 0$$

Testing \( x = 2 \):

$$2^3 + 8(2)^2 + 13(2) - 12 = 8 + 32 + 26 - 12 = 54 \neq 0$$

Testing \( x = -3 \):

$$(-3)^3 + 8(-3)^2 + 13(-3) - 12 = -27 + 72 - 39 - 12 = -6 \neq 0$$

If none of the rational roots work, numerical methods or graphing may be required to find approximate solutions.

Interdisciplinary Connections

The laws of logarithms and logarithmic functions are pivotal in various fields such as physics, engineering, and economics. For instance:

• Physics: The Richter scale for measuring earthquake magnitudes is logarithmic.
• Engineering: Logarithmic scales help in analyzing signal strengths and decibels.
• Economics: Exponential growth models, which rely on logarithms, are used to describe compound interest and population growth.

These applications demonstrate the versatility and importance of mastering logarithmic laws beyond the classroom.

Logarithmic Differentiation

Logarithmic differentiation is a technique used to differentiate functions that are products or quotients of multiple terms or have exponents as functions. By taking the natural logarithm of both sides, the properties of logarithms can simplify differentiation:

For example, to differentiate \( y = x^x \), take the natural logarithm:

$$\ln y = x \ln x$$

Differentiate both sides with respect to \( x \):

$$\frac{1}{y} \frac{dy}{dx} = \ln x + 1$$

Solve for \( \frac{dy}{dx} \):

$$\frac{dy}{dx} = y (\ln x + 1) = x^x (\ln x + 1)$$

This technique is invaluable for handling complex differentiation problems.

Comparison Table

Summary and Key Takeaways