Problem 31 - Algebra 2 and Trigonometry

Question

In \(3-37,\) express each power as a rational number in simplest form. $$ 4^{\frac{1}{3}} \div 4^{\frac{1}{6}} $$

Step-by-Step Solution

Answer

The simplest form of the expression is \( 4^{\frac{1}{6}} \).

Step 1: Apply the Quotient Rule of Exponents

When dividing powers with the same base, we can subtract the exponents: \[ a^m \div a^n = a^{m-n} \] In this problem, we have a base of 4. Therefore, \[ 4^{\frac{1}{3}} \div 4^{\frac{1}{6}} = 4^{\frac{1}{3} - \frac{1}{6}} \] Now, we need to find \( \frac{1}{3} - \frac{1}{6} \).

Step 2: Simplify the Exponents

To subtract the fractions \( \frac{1}{3} - \frac{1}{6} \), we need a common denominator. The least common denominator of 3 and 6 is 6. Convert \( \frac{1}{3} \) to have a denominator of 6:\[ \frac{1}{3} = \frac{2}{6} \]Now subtract the fractions:\[ \frac{2}{6} - \frac{1}{6} = \frac{1}{6} \] So \[ 4^{\frac{1}{3} - \frac{1}{6}} = 4^{\frac{1}{6}} \].

Step 3: Express the Power as a Rational Number

Now we need to express \( 4^{\frac{1}{6}} \) as a rational number. The expression \( 4^{\frac{1}{6}} \) refers to the sixth root of 4. We are not asked to calculate this further since it's not a typical integer or simple fraction, but if decimal approximation or simplification is needed, it can be further evaluated using a calculator. For now, the rational form is \( 4^{\frac{1}{6}} \).

Key Concepts

Quotient Rule of Exponents

The Quotient Rule of Exponents is a fundamental concept in the realm of mathematics, especially when dealing with powers and exponents. This rule simplifies the division of two exponential expressions that have the same base by subtracting their exponents. Here, in our exercise, taking two terms with base 4 allows us to effectively utilize this rule. The general formula is:

• \( a^m \div a^n = a^{m-n} \)

This means, when you divide \( a^m \) by \( a^n \), you are essentially asking, "How many more times has the base been multiplied in the numerator compared to the denominator?" For instance, with our specific problem, \( 4^{\frac{1}{3}} \div 4^{\frac{1}{6}} \), it boils down to subtracting \( \frac{1}{6} \) from \( \frac{1}{3} \). This step simplifies both the computation and understanding of powers by reducing polynomial complexity through direct subtraction.

Fraction Subtraction

Fraction subtraction is often perceived as tedious, but with a systematic approach, it can be quite manageable. The key issue here revolves around finding a common denominator, which serves as a unifier for the subtraction process.For instance, in our exercise, to subtract \( \frac{1}{3} \) from \( \frac{1}{6} \), we require a common denominator. Between 3 and 6, the least common denominator (LCD) is 6. This means we convert \( \frac{1}{3} \) to an equivalent fraction with a denominator of 6:

• \( \frac{1}{3} = \frac{2}{6} \)

With both fractions now having the same denominator, subtraction becomes straightforward:

• \( \frac{2}{6} - \frac{1}{6} = \frac{1}{6} \)

This result effectively tells us the exponent in our problem when two specific powers of 4 are divided.

Simplifying Exponents

Simplifying exponents is about expressing a given power in its simplest rational form. This not only makes complex computations more approachable but also retains the clarity in expressions involving powers or roots.In our scenario, we deduced \( 4^{\frac{1}{6}} \) after performing the subtraction. The expression \( 4^{\frac{1}{6}} \) indicates the sixth root of 4. While the sixth root isn't a neat integer or a simple fraction, it represents the basic idea of a "power" in rational exponent notation. Understanding this representation:

• The numerator \( 1 \) signifies the power of the base (4) raised to a value resulting in the sixth root.
• The denominator \( 6 \) represents dividing or taking a root of the particular power.

While we might not necessarily compute this to a decimal in this exercise, it's essential to perceive the rational exponent form as a minimal expression, providing a neat and tidy way to work algebraically with complex powers without delving into calculators.