Problem 5 - Elementary and Intermediate Algebra

Question

Fill in the blanks. In problem \(6,\) also give the name of each rule. a. \(\log _{b} 1=\) b. \(\log _{b} b=\) c. \(\log _{b} b^{x}=\) d. \(b^{\log _{b} x}=\)

Step-by-Step Solution

Answer

a. 0; b. 1; c. x; d. x.

Step 1: Understanding Logarithm Basics

First, we need to recall some basic properties of logarithms that will help us fill in the blanks. The logarithm function is the inverse of the exponentiation function. Specifically, for a logarithm of the form \( \log_b(x) \), it answers the question: 'To what power must \( b \) be raised, to produce \( x \)?'.

Step 2: Solve Part a

For \( \log_b(1) \), the question asks us for the power to which \( b \) must be raised to equal 1. Any number raised to the power of 0 is 1. Hence, \( \log_b(1) = 0 \). This rule is known as the "Logarithm of 1" rule.

Step 3: Solve Part b

For \( \log_b(b) \), we want to know the power to which \( b \) must be raised to get \( b \) itself. The answer is 1, making \( \log_b(b) = 1 \). This rule is called the "Identity Law of Logarithms."

Step 4: Solve Part c

For \( \log_b(b^x) \), we express \( b^x \) as a power of \( b \), which is already given as \( b^x \). The answer is simply \( x \) since we are asking what power of \( b \) results in \( b^x \). Thus, \( \log_b(b^x) = x \). This rule is the "Power Rule of Logarithms."

Step 5: Solve Part d

For \( b^{\log_b(x)} \), we consider that the logarithm \( \log_b(x) \) returns the power you need to raise \( b \) to get \( x \). Thus multiplying \( b \) by \( \log_b(x) \) results directly in \( x \). Therefore, \( b^{\log_b(x)} = x \). This is the "Inverse Property of Exponents and Logs."

Key Concepts

Logarithm of 1

The concept of the logarithm of 1 is essential to understand the basic properties of logarithms. When we encounter a logarithm in the form \( \log_b(1) \), it is asking us to find the exponent needed to raise the base \( b \) to yield 1.

• The key to remember is that any number raised to the power of 0 equals 1. For example, \( 5^0 = 1 \), \( 10^0 = 1 \), and so on.

Thus, regardless of the base, the logarithm of 1 in any base is 0. In equation form, it is represented as:

• \( \log_b(1) = 0 \)

Identity Law of Logarithms

The identity law of logarithms simplifies expressions where the base and the argument of the logarithm are the same. When you see \( \log_b(b) \), it asks for the power necessary to raise \( b \) to get \( b \).

• This is intuitive because raising a number to the power of 1 results in the number itself.
• For instance, \( 2^1 = 2 \), \( 5^1 = 5 \), and so on.

Therefore, when both base and argument are the same, the logarithm equals 1. Mathematically, this is shown as:

• \( \log_b(b) = 1 \)

Power Rule of Logarithms

The power rule of logarithms is a powerful tool for simplifying logarithmic expressions involving exponents. When looking at an expression like \( \log_b(b^x) \), it asks you to find the exponent that raises base \( b \) to get \( b^x \).

• In this scenario, the power rule directly relates the exponent \( x \) to the equation.
• The base \( b \) raised to the power of \( x \) already yields \( b^x \), so the answer must be \( x \).

This property can be succinctly written as:

• \( \log_b(b^x) = x \)

This rule works because the base on both sides of the expression remains unchanged, simplifying the logarithmic calculation.

Inverse Property of Exponents and Logs

The inverse property of exponents and logarithms illuminates the natural relationship between these mathematical operations. In the expression \( b^{\log_b(x)} \), you're essentially reversing the logarithm through exponentiation.

• The logarithm \( \log_b(x) \) tells us the power needed to raise \( b \) to yield \( x \).
• By taking \( b \) to this power, we naturally return to the original value \( x \).

Hence, the equation is equal to \( x \):

• \( b^{\log_b(x)} = x \)

This property underscores the inverse relationship, allowing for the simplification and solving of logarithmic and exponential expressions.