Problem 55 - Algebra 2 and Trigonometry

Question

In \(27-56,\) evaluate each logarithmic expression. Show all work. $$ \frac{3 \log _{3} 9 \cdot 4 \log _{8} 8 \cdot \log _{13} 169}{6 \log _{2} 256+\log _{\frac{1}{2}} 8} $$

Step-by-Step Solution

Answer

The expression evaluates to \( \frac{16}{15} \).

Step 1: Apply Logarithmic Properties

The problem is to evaluate the expression \( \frac{3 \log_{3} 9 \cdot 4 \log_{8} 8 \cdot \log_{13} 169}{6 \log_{2} 256 + \log_{\frac{1}{2}} 8} \). We start by applying the property \( \log_{b} b^a = a \), which can simplify each logarithmic term. For example, \( \log_{3} 9 = \log_{3} 3^{2} = 2 \).

Step 2: Simplify Numerator

Let's simplify each term in the numerator:- \( \log_{3} 9 = 2 \), so \( 3 \log_{3} 9 = 3 \times 2 = 6 \).- \( \log_{8} 8 = \log_{8} 8^1 = 1 \), so \( 4 \log_{8} 8 = 4 \times 1 = 4 \).- \( \log_{13} 169 = \log_{13} 13^{2} = 2 \).Substitute them back: The numerator becomes \( 6 \times 4 \times 2 = 48 \).

Step 3: Simplify Denominator

Simplify each term in the denominator:- \( \log_{2} 256 = \log_{2} 2^{8} = 8 \), so \( 6 \log_{2} 256 = 6 \times 8 = 48 \).- \( \log_{\frac{1}{2}} 8 \) can be rewritten using \( \log_{\frac{1}{2}} a = -\log_{2} a \): \( \log_{\frac{1}{2}} 8 = -\log_{2} 8 = -3 \).The denominator becomes \( 48 - 3 = 45 \).

Step 4: Calculate the Final Expression

With both the numerator and denominator simplified, we write the expression as \( \frac{48}{45} \). Simplify this fraction by dividing both the numerator and the denominator by their greatest common divisor 3: \( \frac{48 \div 3}{45 \div 3} = \frac{16}{15} \).

Key Concepts

Logarithmic Properties

Understanding logarithmic properties is essential when simplifying logarithmic expressions. A key property often used is \( \log_{b}(b^a) = a \). This tells us that the logarithm of a number to its own base raised to a power is simply that power.
For example:

• \( \log_{3} 9 = \log_{3} (3^2) = 2 \)
• \( \log_{8} 8 = \log_{8} (8^1) = 1 \)
• \( \log_{13} 169 = \log_{13} (13^2) = 2 \)

In each case, the base and the number being logged have a simple relationship, simplifying the logarithmic expression drastically. Knowing these properties helps break down complex logarithmic expressions into manageable steps.

Evaluating Logarithms

When evaluating logarithmic expressions, breaking it down step-by-step is crucial. This means applying known properties to find the numerical value effectively. For example, let's evaluate \( 3 \log_{3} 9 \).
Using the property \( \log_{3} 9 = 2 \), we find the expression becomes \( 3 \times 2 = 6 \). This process also applies to other parts of the expression, like with \( 4 \log_{8} 8 = 4 \times 1 = 4 \) and \( \log_{13} 169 = 2 \).
By calculating each part separately and straightforwardly, you seamlessly reach the evaluation of the entire expression. Always remember to substitute back these calculations into the original expression to ensure accuracy.

Simplifying Logarithmic Expressions

Simplifying logarithmic expressions involves reducing complex forms into simpler ones. Begin by addressing all individual logarithmic terms using known properties. In practice, this often reduces the problem significantly.
Once terms are simplified, like how the numerator becomes \( 6 \times 4 \times 2 = 48 \) and the denominator becomes \( 48 - 3 = 45 \), your task simplifies into basic arithmetic.
Finally, reduce fractions like \( \frac{48}{45} \) by dividing by the greatest common factor, in this case, 3, gives \( \frac{16}{15} \). This simplification process ensures you can handle complex expressions easily and correctly, giving you confidence in your math skills.