Problem 66 - College Algebra with Modeling and Visualization

Question

Exercises \(65-90:\) Use rules of exponents to simplify the expression. Use positive exponents to write your answer. $$ \frac{5 a^{2}}{(x y)^{-1}} $$

Step-by-Step Solution

Answer

The expression simplifies to \( 5a^2xy \) with positive exponents.

Step 1: Understand the Problem

The given expression is \( \frac{5 a^{2}}{(x y)^{-1}} \). We need to simplify it using exponent rules and express the result with positive exponents.

Step 2: Apply Negative Exponent Rule

The rule for negative exponents states \( x^{-n} = \frac{1}{x^n} \). Here, we have \( (xy)^{-1} \), which can be rewritten as \( \frac{1}{xy} \). Thus, the expression becomes \( \frac{5 a^{2}}{\frac{1}{xy}} \).

Step 3: Simplify Division by a Fraction

Dividing by a fraction is equivalent to multiplying by its reciprocal. Therefore, \( \frac{5 a^{2}}{\frac{1}{xy}} \) simplifies to \( 5a^{2} \times xy \).

Step 4: Multiply the Terms

Multiply \( 5a^2 \) by \( xy \). This gives \( 5a^2 \times x \times y = 5a^2xy \).

Step 5: Final Expression with Positive Exponents

The final simplified expression with positive exponents is \( 5a^2xy \).

Key Concepts

Negative Exponents

Negative exponents can be a bit tricky at first, but once you get the hang of them, they'll be a powerful tool in your math toolkit. The rule to remember is: whenever you see a negative exponent, think about flipping the base to turn it positive. For instance, if you encounter an expression like \( x^{-n} \), this is equivalent to \( \frac{1}{x^n} \). Basically, you're taking the reciprocal of the base and getting rid of the negative sign by making the exponent positive.
In our exercise, we found \((xy)^{-1}\). Using our negative exponent rule, we changed this to \(\frac{1}{xy}\). The negative exponent inverts the base, moving it from the denominator to the numerator (or vice versa) as needed. Keeping this rule in mind makes handling negative exponents much simpler.

Simplifying Expressions

Simplifying expressions involves reducing them to their simplest form. This usually means breaking down complex terms and rewriting them to be more manageable. When working with exponents, it's crucial to follow the exponent rules, like the negative exponent rule we discussed.
In our problem, we started with the expression \( \frac{5 a^{2}}{(x y)^{-1}} \). By applying the negative exponent rule, we changed it to \( \frac{5 a^{2}}{\frac{1}{xy}} \). Next, knowing that dividing by a fraction is the same as multiplying by its reciprocal, we turned it into a straightforward multiplication problem: \( 5a^2 \times xy \).
Simplifying expressions often involves transforming complex fractions into easier products, just like we did here. It's all about reshaping the expression in a way that's easier to understand and work with.

Algebraic Multiplication

Algebraic multiplication seems straightforward, but ensuring the correct application of it is vital in obtaining a correct expression. Whether you're multiplying numbers, variables, or a combination of both, keep the multiplication rules in mind to maintain accuracy.
In our exercise, we reached the multiplication part: \( 5a^2 \times xy \). Here, you multiply the constants and variables separately. Start by multiplying the constants (or coefficients) which gives you \(5\) in this case. Then, simply multiply the like bases together under the rules of exponents. This means adding the exponents when the bases are the same.
So \( a^2 \times a^0 = a^{2+0} \), however, since \(xy\) has no exponents shown, it implies each is raised to the power of 1, thus \(x^1y^1\). Finally, assembling these products gives the expression \(5a^2xy\). Remember, clear and careful multiplication keeps expressions accurate and simplified.