Problem 75 - Intermediate Algebra

Question

For Problems \(69-80\), express each of the following as the sum or difference of simpler logarithmic quantities. Assume all variables represent positive real numbers. For example, $$ \begin{aligned} \log _{b} \frac{x^{3}}{y^{2}} &=\log _{b} x^{3}-\log _{b} y^{2} \\ &=3 \log _{b} x-2 \log _{b} y \end{aligned} $$ $$ \log _{b}\left(\frac{x^{\frac{1}{2}} y^{\frac{1}{3}}}{z^{4}}\right) $$

Step-by-Step Solution

Answer

\( \frac{1}{2} \log_b x + \frac{1}{3} \log_b y - 4 \log_b z \)

Step 1: Apply Logarithm Quotient Rule

The given expression is \( \log_b\left( \frac{x^{\frac{1}{2}} y^{\frac{1}{3}}}{z^4} \right) \). Using the quotient rule for logarithms, which states \( \log_b\left( \frac{A}{B} \right) = \log_b A - \log_b B \), we separate the expression into two logarithms. This gives \( \log_b(x^{\frac{1}{2}} y^{\frac{1}{3}}) - \log_b(z^4) \).

Step 2: Apply Logarithm Product Rule

Now, consider the term \( \log_b(x^{\frac{1}{2}} y^{\frac{1}{3}}) \). By the product rule for logarithms, \( \log_b(AB) = \log_b A + \log_b B \), we can split this term into \( \log_b(x^{\frac{1}{2}}) + \log_b(y^{\frac{1}{3}}) \).

Step 3: Apply Power Rule to Simplify Logarithmic Terms

For \( \log_b(x^{\frac{1}{2}}) \) and \( \log_b(y^{\frac{1}{3}}) \), apply the power rule, which states \( \log_b(A^c) = c\log_b A \). This transforms the terms into \( \frac{1}{2}\log_b x \) and \( \frac{1}{3}\log_b y \).

Step 4: Simplify the Final Expression

Substitute back into the expression from Step 1. We have \( \frac{1}{2}\log_b x + \frac{1}{3}\log_b y - 4\log_b z \). Combine all terms to express the original logarithm as the sum and difference of simpler logarithmic quantities.

Key Concepts

Quotient Rule

The Quotient Rule of logarithms is a simple yet powerful tool that helps break down complex fractional expressions into simpler parts. The general formula is \( \log_b\left( \frac{A}{B} \right) = \log_b A - \log_b B \).
This rule tells us that the logarithm of a quotient is equivalent to the difference between the logarithm of the numerator and the logarithm of the denominator.

For example, consider this expression: \( \log_b\left( \frac{x^{\frac{1}{2}} y^{\frac{1}{3}}}{z^4} \right) \). By applying the Quotient Rule, we can rewrite it as:

• \( \log_b(x^{\frac{1}{2}}y^{\frac{1}{3}}) - \log_b(z^4) \)

This step separates the expression into individual logarithmic parts, setting a basis to apply further rules of logarithms. Understanding this separation is crucial for simplifying logarithmic expressions and tackling more complex algebraic structures in later problems.

Product Rule

The Product Rule for logarithms lets us decompose the logarithm of a product into the sum of two individual logarithms. The formula \( \log_b(AB) = \log_b A + \log_b B \) shows that the product inside the logarithm can become a sum outside.

In our ongoing example, we took the expression \( \log_b(x^{\frac{1}{2}} y^{\frac{1}{3}}) \). By applying the Product Rule, it becomes:

• \( \log_b(x^{\frac{1}{2}}) + \log_b(y^{\frac{1}{3}}) \)

This important move leverages the idea that multiplication in the current mathematical landscape of the logarithm is akin to addition. Breaking it down in this way often simplifies understanding and makes the problem more manageable. This approach is especially beneficial when dealing with logarithms of products comprising different bases and powers.

Power Rule

The Power Rule in logarithms takes expressions involving exponentials and expresses them in a linear form. The rule \( \log_b(A^c) = c \log_b A \) emphasizes that you can "bring down" the exponent as a coefficient in front of the logarithm.

Continuing with the example, where we found \( \log_b(x^{\frac{1}{2}}) + \log_b(y^{\frac{1}{3}}) \), the Power Rule allows us to simplify it to:

• \( \frac{1}{2}\log_b x \)
• \( \frac{1}{3}\log_b y \)

This transformation greatly simplifies the expression, turning it into a combination of linear terms. Applying this rule reduces complexity, especially in cases where multiple powers intertwine in logarithmic expressions. Learning to identify powers swiftly inside a logarithm is an effective skill, as this significantly eases the process of rewriting and simplifying potentially cumbersome equations.