Problem 62 - Algebra A Combined Function

Question

If \(\log _{b} 2=0.43\) and \(\log _{b} 3=0.68\), evaluate each expression. $$ \log _{b} \frac{4}{32} $$

Step-by-Step Solution

Answer

\(\log_b \frac{4}{32} = -1.29\)

Step 1: Simplify the Fraction

First, simplify the fraction \( \frac{4}{32} \). Both 4 and 32 can be divided by 4, giving: \( \frac{4}{32} = \frac{1}{8} \).

Step 2: Apply Logarithmic Property

Using the property of logarithms \( \log_b \left( \frac{m}{n} \right) = \log_b m - \log_b n \), apply it to \( \log_b \frac{1}{8} \) as \( \log_b 1 - \log_b 8 \).

Step 3: Evaluate Logarithm of One

Since \( \log_b 1 = 0 \) regardless of the base \( b \), we have \( \log_b 1 = 0 \). Therefore, the expression simplifies to \(- \log_b 8 \).

Step 4: Express 8 as Power of 2

Recognize that \(8 = 2^3 \), so \( \log_b 8 = \log_b (2^3) \).

Step 5: Apply Power Rule for Logarithms

Use the power rule \( \log_b (a^n) = n \cdot \log_b a \). Therefore, \( \log_b (2^3) = 3 \cdot \log_b 2 \).

Step 6: Substitute Given Logarithm Value

Substitute \( \log_b 2 = 0.43 \) into the equation: \( 3 \cdot 0.43 = 1.29 \).

Step 7: Combine Results

Substitute back into the expression from Step 3, \(- \log_b 8 = -1.29 \). Thus, \( \log_b \frac{4}{32} = -1.29 \).

Key Concepts

Logarithmic Properties

Logarithmic properties help us understand how logarithms function and allow us to simplify complex expressions. Some key properties include:

• The Product Rule: This states that \( \log_b (mn) = \log_b m + \log_b n \). It helps in breaking down logarithms with multiplied terms.
• The Quotient Rule: This helps when dealing with divisions and is defined as \( \log_b \left( \frac{m}{n} \right) = \log_b m - \log_b n \).
• The Power Rule: It enables simplification when a number is raised to a power, described as \( \log_b (a^n) = n \cdot \log_b a \).

These properties make evaluating logarithms easier, as they transform complex problems into manageable parts.

Simplifying Expressions

Simplifying expressions involves breaking down mathematical statements into simpler forms. In our exercise, starting with the fraction \( \frac{4}{32} \) simplifies to \( \frac{1}{8} \) by dividing the numerator and the denominator by their greatest common divisor, which is 4.

Next, the logarithmic expression is transformed using the quotient rule: \( \log_b \left( \frac{1}{8} \right) \) becomes \( \log_b 1 - \log_b 8 \). Simplifying expressions like these aids in turning complicated equations into straightforward calculations. The goal is not just to make equations shorter, but also to reveal their structure, making them easier to understand and solve.

Evaluating Logarithms

Evaluating logarithms means finding their values for given numbers. For instance, evaluating \( \log_b \frac{4}{32} \), as shown in the exercise, first involves simplifying the input value into a form suitable for the logarithm properties.

Once simplified to \( \log_b 1 - \log_b 8 \), and knowing that \( \log_b 1 = 0 \) (since any number raised to the power of zero equals 1), we focus on \( \log_b 8 \). This vital step shows how evaluating involves both substituting known values and applying fundamental logarithmic rules to derive new values.

Power Rule for Logarithms

The power rule is a pivotal tool in logarithmic calculations when dealing with expressions involving exponents. According to this rule, \( \log_b (a^n) \) can be expressed as \( n \cdot \log_b a \), effectively pulling down the exponent as a multiplier, which simplifies computation significantly.

In our exercise, recognizing that 8 is \( 2^3 \) allows the application of this rule: \( \log_b (2^3) = 3 \cdot \log_b 2 \). With the known value of \( \log_b 2 = 0.43 \), you can quickly compute that \( 3 \cdot 0.43 = 1.29 \). The power rule thus transforms a potentially overwhelming problem into manageable arithmetic.