Problem 3 - Precalculus

Question

Expand the given logarithm and simplify. Assume when necessary that all quantities represent positive real numbers. $$ \log _{5}\left(\frac{z}{25}\right)^{3} $$

Step-by-Step Solution

Answer

The expanded expression is \( 3 \cdot \log_{5}(z) - 6 \).

Step 1: Apply the Power Rule

The power rule of logarithms states that \( \log_b(a^c) = c \cdot \log_b(a) \). Apply this rule to the given expression: \( \log_{5}\left(\left(\frac{z}{25}\right)^{3}\right) \).This simplifies to: \( 3 \cdot \log_{5}\left(\frac{z}{25}\right) \).

Step 2: Apply the Quotient Rule

The quotient rule of logarithms states that \( \log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y) \). Apply this rule to the expression from Step 1:\( 3 \cdot \left( \log_{5}(z) - \log_{5}(25) \right) \).

Step 3: Simplify the Expression

Now simplify \( \log_{5}(25) \). We know that \( 25 = 5^2 \) so \( \log_{5}(25) = 2 \) because \( \log_{5}(5^2) = 2 \cdot \log_{5}(5) = 2 \).Substitute this into the expression: \( 3 \cdot ( \log_{5}(z) - 2 ) \).

Step 4: Distribute the Constant

Distribute the constant \( 3 \) across the terms inside the parenthesis:\( 3 \cdot \log_{5}(z) - 3 \cdot 2 \).Simplifying gives: \( 3 \cdot \log_{5}(z) - 6 \).

Key Concepts

Power Rule of Logarithms

The power rule of logarithms is a fundamental property that helps us deal with expressions where the argument of the logarithm is raised to a power. It states that for any positive real numbers \( a \) and \( c \), and any positive base \( b \), we have this rule: \[ \log_b (a^c) = c \cdot \log_b (a) \]This means that you can "bring down" the power in front of the logarithm as a multiplier, which significantly simplifies calculations.

• Example: Consider \( \log_{5}(x^3) \). Using the power rule, it's simplified to \( 3 \times \log_{5}(x) \).

When you see a logarithm with an expression raised to a power, remember this rule. It will make the expression easier to work with and understand. In the original problem, we used this rule to rewrite \( \log_{5}\left( \left(\frac{z}{25}\right)^{3}\right) \) as \( 3 \cdot \log_{5}\left(\frac{z}{25}\right) \). This step is crucial for further simplification.

Quotient Rule of Logarithms

The quotient rule of logarithms assists in breaking down logarithms of fractions. This is helpful when the argument of the logarithm is a division of two expressions. According to the quotient rule, for any positive real numbers \( x \) and \( y \), and a positive base \( b \), the rule is:\[ \log_b \left( \frac{x}{y} \right) = \log_b (x) - \log_b (y) \]This rule distributes the logarithm over the division. It effectively separates a single logarithm with a fraction into the difference of two logarithms.

• Example: Consider \( \log_{2}\left(\frac{a}{b}\right) \), it becomes \( \log_{2}(a) - \log_{2}(b) \).

In the context of the original problem, this rule allowed us to take \( \log_{5}(\frac{z}{25}) \) and express it as \( \log_{5}(z) - \log_{5}(25) \). This step laid the groundwork for the entire simplification process.

Logarithmic Simplification

Logarithmic simplification is the process of making a logarithmic expression more concise and easier to understand by reducing its components. The power and quotient rules are valuable tools for achieving this, but the final step often involves simplifying constant values.

• Recognize known values: Simplify logarithmic expressions using known logs. For example, if \( \log_{b}(b^k) \) is encountered, it simply equals \( k \).
• Apply basic arithmetic: Complete any remaining operations such as distributing constants or combining like terms.

In the original problem, after simplifying using the power and quotient rules, we arrived at \( 3 \cdot ( \log_{5}(z) - 2 ) \). By recognizing that \( \log_{5}(25) \) equals 2, we substituted and then distributed \( 3 \) through the expression. The result was a beautifully simplified \( 3 \cdot \log_{5}(z) - 6 \). Understanding and practicing these simplification techniques are vital for mastering logarithms.