Problem 60
Question
Rewrite the expression with positive exponents. $$ (3 x y)^{-2} $$
Step-by-Step Solution
Verified Answer
The expression (3xy)^{-2} rewritten with positive exponents is \(\frac{1}{9x^2y^2}\).
1Step 1: Applying the negative exponent rule
Start by applying the negative exponent rule to the entire expression. Thus, \((3xy)^{-2}\) becomes \(\frac{1}{(3xy)^{2}}\) because of the negative exponent rule.
2Step 2: Squaring the expression in the denominator
The next step will be raising each term in the bracket to the power of two. So, \(\frac{1}{(3xy)^{2}}\) is equivalent to \(\frac{1}{(3^2x^2y^2)}\). This is due to the power of a product rule.
3Step 3: Simplifying the expression
Finally, simplify the expression in the denominator. Hence, the entire expression \(\frac{1}{(3^2x^2y^2)}\) simplifies to \(\frac{1}{9x^2y^2}\).
Key Concepts
Exponent RulesSimplifying ExpressionsPower of a Product
Exponent Rules
Understanding exponent rules is essential when working with mathematical expressions that involve powers. When we talk about exponents, we reference to the number that indicates how many times a base is multiplied by itself. For example, in the expression \(3^2\), the number 3 is the base and 2 is the exponent, indicating that 3 is to be multiplied by itself, thus \(3^2 = 3 \times 3 = 9\).
One of the fundamental exponent rules is the negative exponent rule. It states that any number with a negative exponent is equal to the reciprocal of that number with a positive exponent. For example, \(x^{-a} = \frac{1}{x^a}\). In the exercise, the expression \( (3xy)^{-2} \) immediately invokes this rule, leading us to transform it into a fraction \( \frac{1}{(3xy)^{2}} \) to achieve an equivalent expression with a positive exponent.
Another important rule to remember is the zero exponent rule, which tells us that any nonzero number raised to the power of zero is equal to one, \( x^0 = 1 \) (as long as \( X \) is not zero). These rules are critical for simplifying expressions and solving algebraic problems effectively.
One of the fundamental exponent rules is the negative exponent rule. It states that any number with a negative exponent is equal to the reciprocal of that number with a positive exponent. For example, \(x^{-a} = \frac{1}{x^a}\). In the exercise, the expression \( (3xy)^{-2} \) immediately invokes this rule, leading us to transform it into a fraction \( \frac{1}{(3xy)^{2}} \) to achieve an equivalent expression with a positive exponent.
Another important rule to remember is the zero exponent rule, which tells us that any nonzero number raised to the power of zero is equal to one, \( x^0 = 1 \) (as long as \( X \) is not zero). These rules are critical for simplifying expressions and solving algebraic problems effectively.
Simplifying Expressions
Simplifying expressions is a key skill in algebra that involves rewriting expressions in a simpler or more understandable form without changing their value. The process often involves applying various algebraic rules and properties to reduce complexity and combine like terms.
To simplify a mathematical expression, one must perform operations such as factoring, expanding, combining like terms, and cancelling common factors. For instance, when we see the expression \( \frac{1}{(3xy)^2} \) as in our exercise, our goal is to make it less complex. Squaring each component inside the brackets gives us \( 3^2 \times x^2 \times y^2 \) and simplifying further, it turns into \( \frac{1}{9x^2y^2} \) which is a simpler form.
The exercise demonstrates that identifying and applying the right steps such as squaring the terms individually leads to a more straightforward and often more useful form of the expression. Breaking down the process into clear, manageable parts is an effective way to master simplification.
To simplify a mathematical expression, one must perform operations such as factoring, expanding, combining like terms, and cancelling common factors. For instance, when we see the expression \( \frac{1}{(3xy)^2} \) as in our exercise, our goal is to make it less complex. Squaring each component inside the brackets gives us \( 3^2 \times x^2 \times y^2 \) and simplifying further, it turns into \( \frac{1}{9x^2y^2} \) which is a simpler form.
The exercise demonstrates that identifying and applying the right steps such as squaring the terms individually leads to a more straightforward and often more useful form of the expression. Breaking down the process into clear, manageable parts is an effective way to master simplification.
Power of a Product
The power of a product rule tells us how to handle exponents when they apply to more than one base that is being multiplied. This particular rule states that when a product of bases is raised to an exponent, the exponent applies to each base individually. Mathematically, it's expressed as \( (ab)^n = a^n \times b^n \) for any real numbers \(a\) and \(b\) and an integer \(n\).
In our exercise, we utilize this rule for the term \( (3xy)^2 \) in the denominator. According to the power of a product rule, we square each term in the brackets separately: \(3^2\), \(x^2\), and \(y^2\). This is why we get \( \frac{1}{(3^2 \times x^2 \times y^2)} \) as the next step in the solution process. It's imperative to comprehend that each base \( (3, x, and y) \) is raised to the second power independently, and that's why we later simplify it to \( \frac{1}{9x^2y^2} \) without altering the expression's value.
This rule is extremely helpful in algebra because it allows us to deal with more complicated expressions involving products raised to powers, ensuring that each base is properly accounted for.
In our exercise, we utilize this rule for the term \( (3xy)^2 \) in the denominator. According to the power of a product rule, we square each term in the brackets separately: \(3^2\), \(x^2\), and \(y^2\). This is why we get \( \frac{1}{(3^2 \times x^2 \times y^2)} \) as the next step in the solution process. It's imperative to comprehend that each base \( (3, x, and y) \) is raised to the second power independently, and that's why we later simplify it to \( \frac{1}{9x^2y^2} \) without altering the expression's value.
This rule is extremely helpful in algebra because it allows us to deal with more complicated expressions involving products raised to powers, ensuring that each base is properly accounted for.
Other exercises in this chapter
Problem 59
Solve the equation. Round the result to the nearest hundredth. Check the rounded solution. $$11 k+12=-9$$
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Simplify the expression. \(\left(a b c^{2}\right)^{3} \cdot a b\)
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Evaluate the expression for the given value of the variable(s). $$b^{2}-4 a c when a=1, b=5, c=3$$
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Use the example on the previous page as a model. The average weight \(w\) (in pounds) of an Atlantic cod can be modeled by \(w=1.21(1.42)^{t}\) where \(t\) is t
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