Problem 69 - A Graphical Approach to College Algebra

Question

Use the properties of logarithms to rewrite each logarithm if possible. Assume that all variables represent positive real numbers. $$\log _{5} \frac{5 \sqrt{7}}{3 m}$$

Step-by-Step Solution

Answer

\( 1 + \frac{1}{2} \log_5 7 - \log_5 3 - \log_5 m \)

Step 1: Apply the Quotient Rule

The Quotient Rule of logarithms states that \( \log_b \left( \frac{x}{y} \right) = \log_b x - \log_b y \). We apply this rule to \( \log_5 \left( \frac{5 \sqrt{7}}{3m} \right) \): \[ \log_5 \left( \frac{5 \sqrt{7}}{3m} \right) = \log_5 (5 \sqrt{7}) - \log_5 (3m) \]

Step 2: Apply the Product Rule

The Product Rule of logarithms is \( \log_b (xy) = \log_b x + \log_b y \). We apply it to \( \log_5 (5 \sqrt{7}) \): \[ \log_5 (5 \sqrt{7}) = \log_5 5 + \log_5 \sqrt{7} \]

Step 3: Simplify using the logarithm of a power

The logarithm \( \log_5 5 = 1 \) because any logarithm of its own base is 1. Also, rewrite \( \log_5 \sqrt{7} \) using the power rule: \[ \sqrt{7} = 7^{1/2}, \text{ so } \log_5 \sqrt{7} = \log_5 (7^{1/2}) = \frac{1}{2} \log_5 7 \] Thus, \[ \log_5 (5 \sqrt{7}) = 1 + \frac{1}{2} \log_5 7 \]

Step 4: Apply the Product Rule to the denominator

Apply the Product Rule to the denominator term \( \log_5 (3m) \), breaking it into two separate logarithms: \[ \log_5 (3m) = \log_5 3 + \log_5 m \]

Step 5: Combine all parts

Combine the results from previous steps. Use the equation from Step 1 and substitute the results from Step 2 and Step 4: \[ \log_5 \left( \frac{5 \sqrt{7}}{3m} \right) = (1 + \frac{1}{2} \log_5 7) - (\log_5 3 + \log_5 m) \] Simplifying gives: \[ \log_5 \left( \frac{5 \sqrt{7}}{3m} \right) = 1 + \frac{1}{2} \log_5 7 - \log_5 3 - \log_5 m \]

Step 6: Final expression

The final expression, using the properties of logarithms, is: \[ 1 + \frac{1}{2} \log_5 7 - \log_5 3 - \log_5 m \]

Key Concepts

Quotient Rule

The quotient rule in logarithms helps us decompose a logarithm of a fraction into a subtraction of two logs. It's a valuable tool when you're working with expressions that involve division inside a logarithm. Here's how it works: if you have a logarithm of a fraction like \( \log_b \left( \frac{x}{y} \right) \), you can split it into two parts: \( \log_b x - \log_b y \). In practice, this means you can

• change a complex division problem into simpler subtraction problems,
• making it easier to solve or simplify the expression efficiently.

For example, with our exercise, \( \log_{5} \left( \frac{5 \sqrt{7}}{3m} \right) \) becomes \( \log_{5} (5\sqrt{7}) - \log_{5} (3m) \) using the quotient rule. This step simplifies the problem by separating the numerator and the denominator.

Product Rule

The product rule of logarithms is a handy property for breaking down logs of products into sums of logs. The rule states that for any logarithm \( \log_b (xy) = \log_b x + \log_b y \). This is useful because

• we can break down a complex multiplication inside a logarithm into simpler addition expressions,
• facilitating the step-by-step simplification or calculation process.

Consider the part of our exercise where we deal with \( \log_{5} (5\sqrt{7}) \). By applying the product rule, we rewrite it as \( \log_{5} 5 + \log_{5} \sqrt{7} \). This step makes each individual term easier to handle.

Logarithm Properties

Logarithm properties encompass various rules that help in simplifying and solving logarithmic expressions. Some crucial properties include:

• The identity \( \log_b b = 1 \), where any log that has the same base and argument equals 1.
• Using log transformations for powers and radicals which are explained by the power rule.

In our example, applying these properties reveals that \( \log_{5} 5 = 1 \), which simplifies part of the expression straight away. Utilizing these core properties can significantly streamline the handling of logs.

Logarithm of a Power

Logarithms of powers can be simplified using the power rule. This rule says \( \log_b (x^a) = a \cdot \log_b x \), which means you can take the exponent and place it as a multiplier in front of the logarithm.For instance,

• if you have a square root like \( \sqrt{7} \), this is the same as \( 7^{1/2} \).
• Applying the power rule, \( \log_5 \sqrt{7} \) becomes \( \frac{1}{2} \log_5 7 \).

This step is important as it transforms complex expressions into computationally friendlier terms. Simplifying \( \log_5 (5 \sqrt{7}) \) with \( 1 + \frac{1}{2} \log_5 7 \) shows how breaking down a power log aids in simplification.