Problem 28 - Algebra and Trigonometry

Question

In Exercises 21–42, evaluate each expression without using a calculator. $$ \log _{3} \frac{1}{9} $$

Step-by-Step Solution

Answer

The value of the expression \(\log _{3} \frac{1}{9} = -2\).

Step 1: Understanding the logarithmic equation

Firstly, recognize the operation. It is a logarithm to base 3, noted as \(\log_3\). The expression that needs to be evaluated is \(\log_3 \frac{1}{9}\). According to the formula, this expression is equal to \(n\), where \(3^n = \frac{1}{9}\).

Step 2: Evaluating n

The equation needs to be solved for \(n\). Knowing \(9 = 3^2\), can rewrite the right side of the equation to express 1/9 in terms of 3. Thus, convert 1/9 as \((3^2)^{-1}\) which is equal to \(3^{-2}\). Hence, \(n = -2\).

Key Concepts

Logarithm Properties

Logarithms are mathematical operations that help us understand and solve problems involving exponents. They are the inverses of exponential functions and follow certain properties which make them useful for simplifying complex expressions. There are three key properties of logarithms that are very important:

• The Product Property: This states that the logarithm of a product is the sum of the logarithms of the individual factors. In mathematical terms, \( \log_b(mn) = \log_b(m) + \log_b(n) \).
• The Quotient Property: This tells us that the logarithm of a quotient is the difference of the logarithms of the numerator and the denominator. Formally, \( \log_b\left(\frac{m}{n}\right) = \log_b(m) - \log_b(n) \).
• The Power Property: It dictates that the logarithm of a power is the exponent times the logarithm of the base value. Hence, \( \log_b(m^p) = p \cdot \log_b(m) \).

Using these properties not only simplifies calculations but also plays an integral role in solving logarithmic equations, as in our example from the textbook exercise.

Evaluating Logarithms

Evaluating logarithms means finding the exponent that transforms the base into the desired number. It's essential to understand exactly what a logarithm is asking. For example, when you see \( \log_b(a) \) it is asking, 'To what power must we raise \(b\) to get \(a\)?' This is crucial in solving \( \log_3 \frac{1}{9} \).

The step-by-step solution starts by rewriting the expression with the base \(3\), which is squared to get \(9\) and then inverted to represent \(\frac{1}{9}\). We use the power property of logarithms here, highlighting that \(\log_b\left(m^p\right) = p \cdot \log_b(m)\), to find that \(n\) is the exponent \(-2\). In simpler terms, the number \(-2\) is what \(3\) has to be raised to, to result in \(\frac{1}{9}\). This process of asking and rewriting is a powerful tool to evaluate logarithms.

Exponential Form

The exponential form of a logarithm shows the relationship between logarithms and exponents directly. Think of logarithms as detectives trying to find the missing exponent. Let's say you're given \( \log_b(n) = m \) where \(b\) can't be 1 or 0 since their logs are undefined. This equation tells us that \(b\) raised to \(m\) equals \(n\). So, the exponential form of this logarithmic statement is \(b^m = n\).

In the textbook example, after deducing \(n\) is \(-2\) using logarithmic properties, we confirm our answer because it fits perfectly into the exponential form. We have the equation \(3^n = \frac{1}{9}\text{, and with our calculated value of }n=-2\text{, }\3^{-2} = \frac{1}{9}\), which is exactly the exponential form reflecting our logarithmic expression. Understanding how to flip between these forms is a fundamental skill in algebra that allows students to solve exponential and logarithmic equations effectively.