Problem 22 - Precalculus

Question

Use the properties of logarithms to write the expression as a single logarithm. $$ -\frac{1}{3} \ln (x)-\frac{1}{3} \ln (y)+\frac{1}{3} \ln (z) $$

Step-by-Step Solution

Answer

The expression simplifies to \(\ln\left(\frac{z^{1/3}}{(xy)^{1/3}}\right)\).

Step 1: Use the Logarithm Power Rule

The power rule of logarithms states that \(k \ln(a) = \ln(a^k)\). Apply this to each term in the expression: \(-\frac{1}{3} \ln(x) = \ln(x^{-1/3})\), \(-\frac{1}{3} \ln(y) = \ln(y^{-1/3})\), and \(\frac{1}{3} \ln(z) = \ln(z^{1/3})\).

Step 2: Combine Using Sum of Logarithms

The sum of logarithms property \(\ln(a) + \ln(b) = \ln(ab)\) allows us to combine these logarithms: \(\ln(x^{-1/3}) + \ln(y^{-1/3}) + \ln(z^{1/3}) = \ln(x^{-1/3} y^{-1/3} z^{1/3})\).

Step 3: Simplify the Expression

Recognize that the product inside the logarithm \(x^{-1/3} y^{-1/3} z^{1/3}\) can be rewritten as a single fraction: \(\ln\left(\frac{z^{1/3}}{x^{1/3} y^{1/3}}\right)\). Thus, the entire expression simplifies to a single logarithm, \(\ln\left(\frac{z^{1/3}}{(xy)^{1/3}}\right)\).

Key Concepts

Logarithm Power Rule

The logarithm power rule is a fundamental property that simplifies expressions where a logarithm is multiplied by a constant. If you encounter an expression like \( k \ln(a) \), the power rule allows you to write this as \( \ln(a^k) \). This conversion is particularly handy when you need to manipulate and combine logarithmic expressions.

• For example, \( -\frac{1}{3} \ln(x) \) becomes \( \ln(x^{-1/3}) \).
• It's the same transformation for other variables: \( -\frac{1}{3} \ln(y) = \ln(y^{-1/3}) \) and \( \frac{1}{3} \ln(z) = \ln(z^{1/3}) \).

When applying this rule, you're essentially using the idea that raising a base to a power inside the logarithm is equivalent to a multiplication factor outside it. This is useful in simplifying expressions before combining them.

Sum of Logarithms

The sum of logarithms property is another essential rule for simplifying logarithmic expressions. It states that \( \ln(a) + \ln(b) = \ln(ab) \). This means you can combine separate logarithmic terms into one, as long as they share the same base.
This property greatly simplifies calculations by reducing multiple logarithmic terms into a single one. In the exercise, we applied this property to combine three logarithmic expressions:

• \( \ln(x^{-1/3}) + \ln(y^{-1/3}) + \ln(z^{1/3}) = \ln(x^{-1/3} y^{-1/3} z^{1/3}) \).

By merging the terms in this way, you make the expression easier to handle and prepare it for further simplification.

Single Logarithm

Transforming an expression into a single logarithm emphasizes how the properties of logarithms simplify complex expressions.
Once you've applied both the power rule and the sum of logarithms property, combining them results in a single, tidy logarithmic expression.

• From the exercise, the terms were combined to form \( \ln(x^{-1/3} y^{-1/3} z^{1/3}) \).
• This further simplifies to \( \ln\left( \frac{z^{1/3}}{(xy)^{1/3}} \right) \), showing the power of simplifying under a single logarithm.

Converting a complex expression to a single logarithm not only makes the equation simpler to interpret, but it also offers a clear representation that is often easier to evaluate or further manipulate.

Precalculus

Precalculus serves as the bridge to more advanced mathematics, introducing students to foundational concepts like logarithms. In this journey, students learn to handle and simplify logarithmic expressions—a crucial skill needed for calculus.
Understanding properties such as the logarithm power rule and the sum of logarithms is vital. These properties are not just abstract ideas but practical tools that arise often in algebra, calculus, and beyond.

• Mastering these tools in precalculus ensures smoother progress when tackling calculus topics.
• Precalculus often challenges students to think critically about the symbolic language of mathematics, encouraging precision and analytical skills.

By grasping these concepts early on, you set a solid foundation for future success in advanced math courses.