Problem 87 - Intermediate Algebra

Question

Use rational expressions to write as a single radical expression. $$ \frac{\sqrt[3]{b^{2}}}{\sqrt[4]{b}} $$

Step-by-Step Solution

Answer

The expression simplifies to \( \sqrt[12]{b^5} \).

Step 1: Express each radical with fractional exponents

Convert the expression into fractional exponents. For the cube root, \( \sqrt[3]{b^2} = b^{2/3} \). For the fourth root, \( \sqrt[4]{b} = b^{1/4} \).

Step 2: Write the expression using fractional exponents

The given expression \( \frac{\sqrt[3]{b^{2}}}{\sqrt[4]{b}} \) can now be rewritten using these fractional exponents as \( \frac{b^{2/3}}{b^{1/4}} \).

Step 3: Simplify the expression using the laws of exponents

When dividing powers with the same base, subtract the exponents: \( b^{2/3 - 1/4} \).

Step 4: Find a common denominator

To subtract the exponents \( \frac{2}{3} - \frac{1}{4} \), find a common denominator. The common denominator of 3 and 4 is 12. Rewrite \( \frac{2}{3} \) as \( \frac{8}{12} \) and \( \frac{1}{4} \) as \( \frac{3}{12} \).

Step 5: Subtract the exponents

Subtract the exponents: \( \frac{8}{12} - \frac{3}{12} = \frac{5}{12} \). Thus, the expression becomes \( b^{5/12} \).

Step 6: Convert back to radical notation

Express the result back as a radical. \( b^{5/12} \) can be written as \( \sqrt[12]{b^5} \). Therefore, the expression \( \frac{\sqrt[3]{b^{2}}}{\sqrt[4]{b}} \) is equal to \( \sqrt[12]{b^5} \).

Key Concepts

Fractional Exponents

Fractional exponents are a way to represent roots and powers together in a single expression. Instead of using radical symbols always, fractional exponents offer a concise form. For example, the cube root of a number squared, which is written as \( \sqrt[3]{b^2} \), can be expressed as \( b^{2/3} \). Here, the denominator of the exponent (3) represents the type of root, while the numerator (2) shows the power or the exponent.This concept allows for easier computation as it applies the usual rules of exponents you are familiar with, such as multiplication and division. When you encounter a square root, it can be written as a fractional exponent, \( b^{1/2} \), and similarly, a fourth root becomes \( b^{1/4} \). By transitioning between these forms, they make solving complex radical expressions more manageable.

Laws of Exponents

The laws of exponents summarize how to handle expressions that have bases and exponents. They're particularly useful when dealing with fractional exponents or radicals. A critical law is when dividing two expressions with the same base, we subtract the exponents: \( a^m / a^n = a^{m-n} \).

• Multiplication: When multiplying with the same base, add the exponents, \( a^m \times a^n = a^{m+n} \).
• Power of a Power: Raise a power to a power by multiplying the exponents, \( (a^m)^n = a^{mn} \).
• Negative Exponents: A negative exponent indicates a reciprocal, \( a^{-m} = 1/a^m \).

For the problem at hand, when simplifying \( \frac{b^{2/3}}{b^{1/4}} \), subtract the exponents: \( b^{2/3 - 1/4} \). This gives a simplified form \( b^{5/12} \). Understanding these laws allows you to manipulate and simplify expressions efficiently.

Simplifying Radicals

Simplifying radicals involves rewriting complex root expressions more elegantly or compactly. This often means converting them to fractional exponents and then using laws of exponents for simplification. After computation, you may need to express the solution back in radical form for it to be most intuitive, especially depending on context or instruction. For instance, \( b^{5/12} \) is turned into \( \sqrt[12]{b^5} \). It dictates that you take the twelfth root, simplifying within if possible, and also ending up with a power of five.Here are some steps:

• Convert radicals to fractional exponents for easy manipulation.
• Simplify using the laws of exponents as needed.
• Determine if the radicand can be broken down into simpler factors.
• Convert back to radical format, if required.

Recognizing when to switch between fractional exponents and radicals is key to simplifying expressions elegantly. This dual representation provides flexibility and deeper insight into algebraic problems.