Problem 38 - A Graphical Approach to College Algebra

Question

Perform the indicated operations. Write your answers with only positive exponents. Assume that all variables represent positive real numbers. $$\left(x^{-2} y^{1 / 3}\right)^{5} \cdot\left(8 x^{2} y^{-2}\right)^{-1 / 3}$$

Step-by-Step Solution

Answer

The simplified expression with positive exponents is \(\frac{1}{2} \frac{y^{7/3}}{x^{32/3}}\).

Step 1: Simplify the First Expression

The expression is \((x^{-2} y^{1/3})^{5}\). Apply the power rule for exponents, which states that \((a^m)^n = a^{mn}\). This gives us \(x^{-2 \times 5}\) and \(y^{(1/3) \times 5}\), resulting in \(x^{-10} y^{5/3}\).

Step 2: Simplify the Second Expression

The expression is \((8 x^{2} y^{-2})^{-1/3}\). First, apply the power rule: \((a^m)^n = a^{mn}\). Simplify each part: \(8^{-1/3}\), \(x^{2 \times -1/3}\), and \(y^{-2 \times -1/3}\). This results in \(8^{-1/3}\), \(x^{-2/3}\), and \(y^{2/3}\).

Step 3: Simplify 8^{-1/3}

The expression \(8^{-1/3}\) can be simplified using cube roots. Since \(8 = 2^3\), we have \(8^{-1/3} = (2^3)^{-1/3} = 2^{-1} = \frac{1}{2}\).

Step 4: Combine the Simplified Expressions

Now, multiply the results from Steps 1 and 2: \(x^{-10} y^{5/3} \times \frac{1}{2} x^{-2/3} y^{2/3}\). Simplify by adding the exponents for like bases: \(x^{-10 + (-2/3)}\) and \(y^{5/3 + 2/3}\). Also, multiply the coefficients: \(1 \times \frac{1}{2} = \frac{1}{2}\).

Step 5: Simplify Like Terms

Calculate \(x^{-10 + (-2/3)}\) which translates to \(x^{-(10 + 2/3)} = x^{-32/3}\) and \(y^{5/3 + 2/3} = y^{7/3}\). Combine with the coefficient to obtain \(\frac{1}{2} x^{-32/3} y^{7/3}\).

Step 6: Convert Negative Exponents to Positive

Rewrite the expression with positive exponents: \(\frac{1}{2} \frac{y^{7/3}}{x^{32/3}}\). This is achieved by moving \(x^{-32/3}\) to the denominator.

Key Concepts

Power Rule for Exponents

The power rule for exponents is a fundamental principle in mathematics that helps simplify expressions involving powers. It states that when an exponent is raised to another exponent, you multiply the exponents: \[(a^m)^n = a^{m \times n}\].
This rule allows you to transform complex expressions into more manageable forms. For instance, if you have \((x^{-2} y^{1/3})^5\), applying the power rule gives \(x^{-2 \times 5}\) and \(y^{1/3 \times 5}\).
This results in \(x^{-10}\) and \(y^{5/3}\). Breaking down problems this way simplifies the process, making them easier to solve. Remember: always multiply the exponents and keep the base the same. This makes the seemingly complex task of exponentiation straightforward and efficient.

Simplifying Expressions

Simplifying expressions involves combining like terms and reducing the expression to its simplest form. In our problem, we simplify by looking at like bases and combining their exponents. For example, multiplying \(x^{-10} y^{5/3}\) by \(\frac{1}{2} x^{-2/3} y^{2/3} \) involves:

• Adding together the exponents of similar bases, like \(x\) and \(y\).
• Calculating the sum of exponents: \(x^{-10 + (-2/3)} = x^{-32/3}\) and \(y^{5/3 + 2/3} = y^{7/3}\).
• Multiplying the coefficients: \(1 \times \frac{1}{2} = \frac{1}{2}\).

This process transforms the expression into a simplified form: \(\frac{1}{2} x^{-32/3} y^{7/3}\). Simplifying expressions makes complex calculations more accessible by breaking them into smaller, more manageable parts.

Cube Roots

Cube roots are the inverse operation of raising a number to the power of three. They help simplify expressions and are particularly useful in problems involving exponents.
To find the cube root, you essentially ask yourself, "What number, when multiplied by itself three times, gives the original number?" For example, \(8\) is equal to \(2^3\), so the cube root of 8 is 2.
In our expression, we deal with \(8^{-1/3}\). Since the cube root of \(8\) is \(2\), the expression becomes \((2^3)^{-1/3} = 2^{-1} = \frac{1}{2}\). This is crucial for simplifying your calculations by converting complex numbers into more recognizable forms. Understanding cube roots aids in breaking down and solving exponentiation tasks effectively.