Problem 80 - Precalculus

Question

$$\text { Show that } \log _{b}\left(\frac{1}{x}\right)=-\log _{b} x$$

Step-by-Step Solution

Answer

The equation is true by using the property \( \log_{b}(\frac{1}{x}) = -\log_{b} x \).

Step 1: Apply Logarithmic Property

We start by using the logarithmic property: the logarithm of a quotient can be written as the subtraction of the logarithms. For the logarithm in the statement, we have: \[ ext{log}_{b} rac{1}{x} = ext{log}_{b}(1) - ext{log}_{b} x \]

Step 2: Evaluate Logarithm of 1

The logarithm of 1 with any base is 0. Thus, we substitute \( \text{log}_{b}(1) = 0 \) into the equation:\[ ext{log}_{b} rac{1}{x} = 0 - ext{log}_{b} x \]

Step 3: Simplify the Expression

Simplify the equation from the previous step:\[ ext{log}_{b} rac{1}{x} = - ext{log}_{b} x \]This shows that the original statement \( \text{log}_{b}igg(\frac{1}{x}\bigg) = -\text{log}_{b} x \) is true.

Key Concepts

Logarithm of a Quotient

The logarithm of a quotient is a fundamental identity in logarithmic mathematics. To put it simply, when you take the logarithm of one number divided by another, you are essentially calculating the difference between the logarithms of these two numbers.
This idea can be represented as:

• \( \log_b\left(\frac{M}{N}\right) = \log_b M - \log_b N \)

This property is incredibly useful for simplifying expressions that involve division within logarithms.
For example, if you have \( \log_b\left(\frac{1}{x}\right) \), you can separate it as \( \log_b 1 - \log_b x \).
In essence, whenever you see a quotient inside a logarithm, you can rewrite it as subtraction between two separate logarithmic expressions. Understanding this property allows for easier manipulation and simplification of complex logarithmic expressions.

Logarithm of 1

The logarithm of 1 is a unique and simple concept. Regardless of the base \( b \) you choose (as long as \( b > 0 \) and \( b eq 1 \)), the logarithm of 1 is always 0.
This is because any number raised to the power of 0 equals 1:

• \( b^0 = 1 \)

So, when you evaluate \( \log_b 1 \), you're essentially asking "To what power must \( b \) be raised to yield 1?". The answer is always 0.
This concept is particularly useful in simplifying expressions. For example, when using the Logarithm of a Quotient identity, you might come across \( \log_b 1 \), which simplifies instantly to 0, making lengthy calculations much easier.
Remembering that \( \log_b 1 = 0 \) is a key tool in simplifying and solving logarithmic expressions.

Properties of Logarithms

Properties of logarithms are mathematical rules that make working with logarithms easier and more intuitive.
These properties are derived from fundamental rules of exponents and can greatly simplify complex logarithmic expressions. Here are some vital properties to know:

• Product Property: \( \log_b (MN) = \log_b M + \log_b N \)
• Quotient Property: \( \log_b \left(\frac{M}{N}\right) = \log_b M - \log_b N \)
• Power Property: \( \log_b (M^n) = n \cdot \log_b M \)
• Logarithm of 1: \( \log_b 1 = 0 \)
• Change of Base Formula: \( \log_b a = \frac{\log_k a}{\log_k b} \), where \( k \) is any positive number

Using these properties, students can break down complex logarithmic problems into simpler forms, making them easier to solve. Whether you are multiplying, dividing, or raising numbers to a power, these properties come in handy to navigate through logarithmic expressions with ease.