Problem 30 - Precalculus : Building Concepts and Connections

Question

Evaluate each expression without using a calculator. $$\log _{7} \frac{1}{49}$$

Step-by-Step Solution

Answer

The value of \( \log _{7} \frac{1}{49} \) is -2

Step 1: Rewrite the argument as a power of 7

Looking at the argument \( \frac{1}{49} \), it can be rewritten as \( 7^{-2} \) because \( 49 = 7^2 \) and \( \frac{1}{7^{2}} = 7^{-2} \)

Step 2: Apply the basic rule of logarithms

Now apply the rule \( \log_b (b^x) = x \). Here \( b = 7 \) and \( x = -2 \). Therefore, \( \log _{7} 7^{-2} = -2 \)

Key Concepts

Logarithmic Expressions

Understanding logarithmic expressions is foundational for solving logarithm-related problems. A logarithm, in its basic form, is the inverse of exponentiation. It answers the question: 'To what exponent must we raise the base to get a certain number?' The general form is written as \(\text{log}_b(a)\), where 'b' is the base and 'a' is the argument of the logarithm.

When evaluating a logarithmic expression like \(\text{log}_7 \frac{1}{49}\), we seek to rewrite the argument in a way that makes the base evident. In this case, recognizing that 49 is a power of 7 assists us in transforming the argument into a power of the base. Specifically, we identify that \(49 = 7^2\), which allows us to express \( \frac{1}{49} \) as \(7^{-2}\).

This transformation is essential because logarithmic functions are designed to unpack exponents, and having the argument as a power of the base simplifies the evaluation of the logarithmic expression considerably.

Logarithm Rules

To effectively evaluate logarithms, one must be familiar with a handful of fundamental rules that govern their behavior. These rules help simplify and solve logarithmic expressions. Among them, key rules include the following:

• Product Rule: \(\text{log}_b(mn) = \text{log}_b(m) + \text{log}_b(n)\)
• Quotient Rule: \(\text{log}_b\bigg(\frac{m}{n}\bigg) = \text{log}_b(m) - \text{log}_b(n)\)
• Power Rule: \(\text{log}_b(m^k) = k \text{log}_b(m)\)
• Change of Base Rule: \(\text{log}_b(m) = \frac{\text{log}_k(m)}{\text{log}_k(b)}\), where 'k' is any positive real number.
• Logarithm of One: \(\text{log}_b(1) = 0\)
• Logarithm of the Base: \(\text{log}_b(b) = 1\)

In the given exercise, we apply the basic rule that states \(\text{log}_b(b^x) = x\). This allows us to directly read off the exponent as the output of the logarithmic expression when the argument is an exponent of the base.

Exponentiation

Exponentiation is a mathematical operation involving two numbers, the base and the exponent. The base 'b' raised to the power of an exponent 'n' is written as \(b^n\), meaning the base is multiplied by itself 'n' times. In the context of logarithms, exponentiation is inversely related, as logarithms help to 'decompose' the exponentiation process.

When we have a power such as \(7^{-2}\), it implies a reciprocal because negative exponents indicate division by that base raised to the positive exponent: \(7^{-2} = \frac{1}{7^2} = \frac{1}{49}\). This is an essential concept because, in our exercise, we deal with \(\text{log}_7\) of a negative exponent.

Understanding how to work with exponents, especially when it comes to negative exponents, plays a crucial role in simplifying logarithms. This knowledge leads to a seamless transition from the expression \(\text{log}_7 \frac{1}{49}\) to recognizing it as \(\text{log}_7 7^{-2}\) and ultimately to evaluating the logarithm, obtaining the exponent as the result.