Problem 31 - Precalculus

Question

Evaluate the expression. \(\log _{36}(\sqrt[4]{36})\)

Step-by-Step Solution

Answer

The value of the expression is \(\frac{1}{4}\).

Step 1: Simplify the Radicand

First, recognize that \(\sqrt[4]{36}\\) can also be expressed as \(36^{1/4}\). By making this substitution, the expression becomes \(\log_{36} (36^{1/4})\).

Step 2: Apply the Power Rule for Logarithms

The power rule for logarithms, \(\log_b (a^n) = n \log_b a\), allows us to move the exponent to the front. Applying this rule, we get \(\frac{1}{4} \cdot \log_{36} 36\).

Step 3: Evaluate the Logarithm

Recognize that \(\log_{b} b = 1\). Since 36 is the base of the logarithm and also the argument in the logarithm function, \(\log_{36} 36\) simplifies to \(1\).

Step 4: Final Calculation

Substituting the result from Step 3 into the expression from Step 2, we get \(\frac{1}{4} \cdot 1 = \frac{1}{4}\). Thus, the simplified value of the expression is \(\frac{1}{4}\).

Key Concepts

Properties of Logarithms

Logarithms are mathematical functions that help us solve problems involving exponents. They come with a set of properties that make calculations easier, especially when simplifying expressions. These properties include the product rule, quotient rule, and power rule. In this exercise, we focus on the power rule, but it's essential to know these others:

• Product Rule: This helps when multiplying numbers inside a logarithm. It states that \( \log_b (mn) = \log_b m + \log_b n \).
• Quotient Rule: Useful for division inside logs: \( \log_b \left(\frac{m}{n}\right) = \log_b m - \log_b n \).
• Power Rule: The one we use here allows us to move an exponent out in front: \( \log_b (a^n) = n \cdot \log_b a \).

Recognizing these properties makes solving equations quicker, as they help simplify complex logarithmic expressions.

Evaluating Expressions

Evaluating expressions with logarithms can feel complicated, but breaking them down into steps simplifies the process. Let's walk through our specific example: evaluating \( \log_{36}(\sqrt[4]{36}) \).First, notice that \( \sqrt[4]{36} \) is equivalent to saying \( 36^{1/4} \). Understanding expressions in fractional exponent form is key, as it prepares them for applying logarithmic properties.Next, apply the power rule. Here, the expression inside the log is \( 36^{1/4} \), so you take the 1/4 exponent and bring it in front of the log: \( \frac{1}{4} \cdot \log_{36} 36 \). This simplification makes the expression much easier to handle.Finally, we use the property that \( \log_b b = 1 \). Since \( 36 \) is both the base and the number in our logarithm, \( \log_{36} 36 \) simplifies to 1, leading to our final result of \( \frac{1}{4} \). Breaking down each step clears up confusion and makes the process manageable.

Power Rule for Logarithms

The power rule is a powerful tool when working with logarithms and exponents. It specifically deals with expressions where the argument of the log is raised to a power. The rule states: \( \log_b (a^n) = n \cdot \log_b a \). By using the power rule, you can simplify the computation significantly. Instead of dealing with complex exponentials, you merely need to move the exponent in front of the log and multiply it by the logarithm itself. For instance, in \( \log_{36}(36^{1/4}) \), the rule lets us write the entire log as \( \frac{1}{4} \cdot \log_{36} 36 \). This makes evaluating the expression straightforward.This rule works because logarithms and exponents are inverse operations. The exponent and the log 'cancel' each other out, reducing the expression to its core components. Understanding the power rule and how to use it simplifies logarithms' complexity and reduces potential mistakes when handling logs with exponents.