Problem 57 - Algebra and Trigonometry

Question

Simplify the expression and eliminate any negative exponents \((\mathrm{s}) .\) Assume that all letters denote positive numbers. \(\left(\frac{3 a^{-2}}{4 b^{-1 / 3}}\right)^{-1}\)

Step-by-Step Solution

Answer

The simplified expression is \(\frac{4 b^{1/3}}{3 a^2}\).

Step 1: Apply the Negative Exponent Rule

The expression given is \(\left(\frac{3 a^{-2}}{4 b^{-1 / 3}}\right)^{-1}\). According to the negative exponent rule, \(x^{-n} = \frac{1}{x^n}\). Thus, applying this rule to the whole expression, we have: \(\left(\frac{3 a^{-2}}{4 b^{-1 / 3}}\right)^{-1} = \frac{4 b^{-1/3}}{3 a^{-2}}\).

Step 2: Simplify the Fraction by Eliminating Negative Exponents

Now, we will simplify \(\frac{4 b^{-1/3}}{3 a^{-2}}\) by addressing the negative exponents. The negative exponents in the expression can be rewritten using the rule \(x^{-n} = \frac{1}{x^n}\). Thus: \(b^{-1/3} = \frac{1}{b^{1/3}}\) and \(a^{-2} = \frac{1}{a^2}\). Substituting these, the expression becomes: \(\frac{4}{3} \cdot \frac{1/a^2}{1/b^{1/3}} = \frac{4 b^{1/3}}{3 a^2}\).

Step 3: Final Expression

After simplifying, the final expression is \(\frac{4 b^{1/3}}{3 a^2}\). All negative exponents have been eliminated, and the expression is fully simplified.

Key Concepts

Negative Exponent Rule

Understanding the negative exponent rule is key to manipulating algebraic expressions with ease. This rule states that any base raised to a negative exponent can be expressed as the reciprocal of that base with a positive exponent. In mathematical terms, this can be represented as:

• If you have a term like \(x^{-n}\), it can be rewritten as \(\frac{1}{x^n}\).
• Similarly, \(\frac{1}{x^{-n}}\) becomes \(x^n\).

This transformation helps simplify complex expressions, making further calculations or simplifications much clearer.
In the given exercise, the expression \(\left(\frac{3 a^{-2}}{4 b^{-1 / 3}}\right)^{-1}\) was simplified first by applying this rule, leading to \(\frac{4 b^{-1/3}}{3 a^{-2}}\).Using the negative exponent rule ensures that expressions become easier to manage, especially when multiple fractions or terms are involved.

Simplifying Expressions

Simplifying expressions is a process that helps in breaking down complex algebraic forms into their most manageable state. Doing so not only aids in understanding but also makes further calculations more straightforward. Let’s consider our expression after applying the negative exponent rule: \(\frac{4 b^{-1/3}}{3 a^{-2}}\).To simplify further, we need to **eliminate** any negative exponents by using the reciprocal properties from the earlier step:

• \(b^{-1/3}\) becomes \(\frac{1}{b^{1/3}}\).
• Similarly, \(a^{-2}\) becomes \(\frac{1}{a^2}\).

Replacing each term, the expression transforms to \(\frac{4}{3} \cdot \frac{1/a^2}{1/b^{1/3}} = \frac{4 b^{1/3}}{3 a^2}\).
This rearrangement simplifies the fraction into a form that's intuitive and user-friendly.By practicing these steps, students can confidently tackle even the most daunting expressions, knowing they can simplify them to their core components.

Algebraic Fractions

Algebraic fractions are similar to numerical fractions but consist of algebraic expressions in the numerator and/or the denominator. These can sometimes involve expressions with exponents, requiring special attention.
For example, the fraction \(\frac{3 a^{-2}}{4 b^{-1 / 3}}\) in our original problem dealt with both negative exponents and algebraic terms.Key steps for handling these include:

• Utilizing the negative exponent rule to turn complex expressions into simpler, reciprocal terms.
• Simplifying by eliminating negative exponents through mutual multiplication or division, where applicable.
• Cleansing the fraction of any unnecessary complexity to achieve the most reduced form.

The goal is to bring the expression into a state where calculation becomes intuitive. For instance, once transformed and simplified, our fraction \(\frac{4 b^{1/3}}{3 a^2}\) reveals clearly the simplified relationship between the terms.
Consistent practice with such problems enhances proficiency in recognizing algebraic structures, leading to faster and more accurate simplifications.