Problem 19 - Algebra 2

Question

Write each expression in radical form. $$ c^{\frac{2}{5}} $$

Step-by-Step Solution

Answer

The expression in radical form is \( \sqrt[5]{c^2} \).

Step 1: Understanding Exponents

The expression given is \( c^{\frac{2}{5}} \). This represents an exponent with a fractional value. In general, \( a^{\frac{m}{n}} \) can be rewritten in terms of a radical and a power.

Step 2: Using the Radical Form

The expression \( c^{\frac{2}{5}} \) can be rewritten in radical form by recognizing the numerator as the power and the denominator as the root. In this case, \( 2 \) is the power and \( 5 \) is the root, which corresponds to the fifth root of \( c \), raised to the power of \( 2 \).

Step 3: Writing the Expression in Radical Form

The fifth root of \( c \) is written as \( \sqrt[5]{c} \). Raising this expression to the power of \( 2 \) gives us \( (\sqrt[5]{c})^2 \). Therefore, \( c^{\frac{2}{5}} \) can be expressed in radical form as \( \sqrt[5]{c^2} \).

Key Concepts

Fractional Exponents

Fractional exponents can seem tricky at first, but they're really just another way to express roots and powers in mathematical expressions. When you see an exponent like \( c^{\frac{2}{5}} \), it's important to understand it represents both a root and a power. This type of exponent is known as a fractional exponent, where the numerator is the power and the denominator is the root.

• The numerator (in this case, \( 2 \)) tells us how many times to multiply the base by itself.
• The denominator (\( 5 \)) indicates the root to be taken of the base.

So, \( c^{\frac{2}{5}} \) tells us that you both find the fifth root of \( c \) and then square (raise to the power of 2) this result. This understanding lays the groundwork for converting into radical form.

Radical Form

Radical form offers a more intuitive way to work with fractional exponents. It connects the ideas of roots and powers directly to the expression. When converting from a fractional exponent like \( c^{\frac{2}{5}} \) to radical form, you'll follow these steps:

• Identify the root based on the denominator of the exponent. Here, the denominator is \( 5 \), so we deal with the fifth root.
• Write this as \( \sqrt[5]{c} \).
• The numerator \( 2 \) represents the power, so you raise the expression to this power, resulting in \( (\sqrt[5]{c})^2 \).

Thus, expressing \( c^{\frac{2}{5}} \) in radical form becomes \( \sqrt[5]{c^2} \). This method is a powerful conversion tool, simplifying complex expressions.

Exponentiation

Exponentiation refers to the process of raising a number to a power. It's a fundamental operation in algebra that allows you to express repeated multiplication in a concise form. When dealing with fractional exponents such as \( c^{\frac{2}{5}} \), you're extending this idea to include roots alongside powers.

• Think of the base \( c \) being multiplied by itself as many times as indicated by the numerator (\( 2 \)).
• Also, the expression involves taking a root because of the denominator (\( 5 \) signifies taking a fifth root).

This dual action of exponentiation with fractional exponents reflects a deeper understanding of how numbers can be manipulated, enhancing your mathematical reasoning skills.

Roots and Powers

The concept of roots and powers is central to understanding radical expressions and exponents. While powers indicate multiplication, roots indicate division in a sense. Specifically, finding the fifth root of a number means determining what number, multiplied by itself five times, gives the original number.

• Powers, such as the \( 2 \) in \( \frac{2}{5} \), simply mean multiplying a number by itself.
• Roots, like the fifth root, require identifying the base that yields the original number when used as a factor that many times.

In the case of \( c^{\frac{2}{5}} \), applying both a root and a power combines these two operations:1. Find the base that is the fifth root of \( c \).2. Multiply this result by itself \( 2 \) times.By understanding these processes, you'll be equipped to tackle any problem involving radical expressions.