Problem 31 - College Algebra with Corequisite Support

Question

For the following exercises, use properties of logarithms to evaluate without using a calculator. \(6 \log _{8}(2)+\frac{\log _{8}(64)}{3 \log _{8}(4)}\)

Step-by-Step Solution

Answer

The evaluated expression is 3.

Step 1: Apply Power Rule

The power rule of logarithms states that \( a \log_b(x) = \log_b(x^a) \). We apply this rule to the first term, \(6 \log _{8}(2)\), so it becomes \(\log _{8}(2^6)\). This simplifies to \(\log _{8}(64)\) since \(2^6 = 64\).

Step 2: Simplify the Numerator

The numerator of the second fraction is already \(\log _{8}(64)\). Therefore, there is no need for simplification or alteration to this part at this point.

Step 3: Simplify the Denominator

The denominator of the second fraction is \(3 \log _{8}(4)\). Using the power rule, we convert this to \(\log _{8}(4^3)\). Since \(4^3 = 64\), it simplifies to \(\log _{8}(64)\).

Step 4: Combine Logarithms

Now, our expression is \(\log _{8}(64) + \frac{\log _{8}(64)}{\log _{8}(64)}\). The fraction simplifies to 1 because \(\frac{\log _{8}(64)}{\log _{8}(64)} = 1\).

Step 5: Finish the Calculation

Combining the simplified terms gives \(\log _{8}(64) + 1\). Now, recognize that since \(8^2 = 64\), \(\log _{8}(64) = 2\). Thus, the expression simplifies to \(2 + 1 = 3\).

Key Concepts

Logarithm Properties

Logarithm properties are valuable tools that help us evaluate and simplify expressions involving logarithms without a calculator. Understanding these properties can make solving such problems much more straightforward. Among the most commonly used properties are:

• 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: \( a \log_b(x) = \log_b(x^a) \)

These properties allow us to break down complex logarithmic expressions into simpler parts. By applying them, we can often transform challenging problems into much easier ones. This exercise, for instance, makes extensive use of the power rule to combine terms and simplify calculations. Becoming familiar with these properties will make working with logarithms a breeze.

Power Rule of Logarithms

The power rule of logarithms is a versatile tool that simplifies expressions with exponential elements. This rule states that if you have a term of the form \( a \log_b(x) \), it can be rewritten as \( \log_b(x^a) \). This transformation is particularly useful when dealing with terms that involve multiplication of a logarithm by a number.Understanding and applying the power rule:

• Take the coefficient outside the logarithm and turn it into an exponent of the term inside the logarithm.
• For example, transform \( 6 \log_8(2) \) into \( \log_8(2^6) \), which simplifies to \( \log_8(64) \).

The power rule essentially allows you to condense the expression, turning multiplication into exponentiation, which can often make the remaining calculations more straightforward. In the given exercise, applying this rule simplifies the terms drastically, helping to evaluate the entire expression more easily.

Simplifying Logarithmic Expressions

Simplifying logarithmic expressions involves breaking down complex structures using various logarithmic properties. The goal is to combine or simplify terms to reach a more elementary form. This exercise showcases this process effectively by:

• Using the power rule to express terms in exponential form.
• Simplifying fractions within the expression by recognizing and canceling terms.
• Combining the simplified terms into a cohesive result.

In the example expression, terms such as \( \log_8(64) \) appear repeatedly. Recognizing this allows for further simplification. Additionally, the fraction \( \frac{\log_8(64)}{\log_8(64)} \) simplifies to 1, reducing complexity in the expression.
Through methodically applying properties like the power rule, and simplifying relations within the expression, it ultimately simplifies to a basic arithmetic calculation: converting \( \log_8(64) \) to 2 (since \( 8^2 = 64 \)), and adding the 1, to achieve a final result of 3. Mastering the simplification process can make working with logarithms less intimidating and more intuitive.