Problem 76 - A Graphical Approach to Precalculus with Limits

Question

Use the properties of logarithms to rewrite each logarithm if possible. Assume that all variables represent positive real numbers. $$\log _{p} \sqrt[3]{\frac{m^{5}}{k t^{2}}}$$

Step-by-Step Solution

Answer

\( \frac{5}{3} \cdot \log_{p}(m) - \frac{1}{3} \cdot \log_{p}(k) - \frac{2}{3} \cdot \log_{p}(t) \)

Step 1: Rewrite the Radical Using Exponents

The expression inside the logarithm is \( \sqrt[3]{\frac{m^{5}}{kt^{2}}} \), which can be rewritten using exponents as \( \left( \frac{m^{5}}{kt^{2}} \right)^{\frac{1}{3}} \). This step helps in simplifying the expression using properties of exponents.

Step 2: Use the Power Rule of Logarithms

Apply the power rule, which states \( \log_{b}(x^a) = a \cdot \log_{b}(x) \). So, \( \log_{p} \left( \frac{m^{5}}{kt^{2}} \right)^{\frac{1}{3}} = \frac{1}{3} \cdot \log_{p} \left( \frac{m^{5}}{kt^{2}} \right) \).

Step 3: Apply the Quotient Rule of Logarithms

The quotient rule for logarithms states \( \log_{b} \left( \frac{x}{y} \right) = \log_{b}(x) - \log_{b}(y) \). Apply this to get \( \log_{p}(m^{5}) - \log_{p}(kt^{2}) \).

Step 4: Apply the Power Rule Again

Use the power rule on each term: \( \log_{p}(m^{5}) = 5 \cdot \log_{p}(m) \) and \( \log_{p}(kt^{2}) = \log_{p}(k) + \log_{p}(t^{2}) = \log_{p}(k) + 2 \cdot \log_{p}(t) \).

Step 5: Combine All Parts Together

Substitute back into the original expression: \( \frac{1}{3} \cdot \left( 5 \cdot \log_{p}(m) - (\log_{p}(k) + 2 \cdot \log_{p}(t)) \right) \). Simplify further to get \( \frac{1}{3} \left( 5 \cdot \log_{p}(m) - \log_{p}(k) - 2 \cdot \log_{p}(t) \right) \).

Key Concepts

Power Rule

The Power Rule is a significant property of logarithms that makes simplifying expressions a breeze. It states that if you have a logarithm of the form \( \log_{b}(x^a) \), it can be rewritten as \( a \cdot \log_{b}(x) \). This rule simplifies the process of managing exponents within a logarithm.
For example, in the expression \( \log_{p} \left( \frac{m^{5}}{kt^{2}} \right)^{\frac{1}{3}} \), the \( \frac{1}{3} \) suggests that the entire fraction is raised to the power of \( \frac{1}{3}\).

**Benefits of the Power Rule**

• Simplification: It breaks down complex expressions, making them easier to handle.
• Efficiency: Calculations can be performed more swiftly, especially when dealing with high powers.

When applying the Power Rule, always start by determining the power of the expression within the logarithm. Then, simply bring the exponent in front of the logarithm, transforming it into a coefficient.

Quotient Rule

The Quotient Rule of logarithms is invaluable when dealing with fractions inside a logarithmic expression. It posits that \( \log_{b} \left( \frac{x}{y} \right) = \log_{b}(x) - \log_{b}(y) \). This rule helps break down the fraction into manageable parts.
Continuing with the example \( \log_{p} \left( \frac{m^{5}}{kt^{2}} \right) \), using the quotient rule simplifies it to \( \log_{p}(m^{5}) - \log_{p}(kt^{2}) \).

**Key Points about the Quotient Rule**

• Decomposition: Convert complex expressions into simpler components.
• Problem Solving: Makes finding solutions more accessible and less daunting.

Utilize the quotient rule when you face division inside a logarithmic expression. Doing so will make it easier to analyze each part separately.

Exponents

Exponents are an essential part of mathematics and are often intertwined with logarithms. They represent the power to which a number or expression is raised. For instance, \( m^{5} \) means that \( m \) is multiplied by itself five times.

In the context of our example, exponents play a vital role when expressing roots as fractional exponents. The expression \( \sqrt[3]{\frac{m^{5}}{kt^{2}}} \) is first converted to \( \left( \frac{m^{5}}{kt^{2}} \right)^{\frac{1}{3}} \) using the properties of exponents.

**Features of Exponents**

• Simplifying Roots: Transforms root expressions into fractional powers, easing calculations.
• Foundational for Logarithms: Often simplified further using the Power Rule.

Whether dealing with whole numbers or fractions, understanding how exponents work is crucial when manipulating mathematical expressions. They act as the backbone for many logarithmic operations.