Problem 34
Question
Simplify the expression, writing your answer using positive exponents only. $$ \left(\frac{a^{-2}}{2 b^{2}}\right)^{-3} $$
Step-by-Step Solution
Verified Answer
\(\frac{b^{6}}{8 a^{6}}\)
1Step 1: Apply the power to each term
Apply the power of -3 to each term inside the parentheses:
$$
\left(\frac{a^{-2}}{2 b^{2}}\right)^{-3} = \frac{a^{-6}}{8 b^{-6}}
$$
We raised each term to the power of -3.
2Step 2: Rewriting with positive exponents
Rewrite the expression using positive exponents by moving terms with negative exponents from the numerator to the denominator and vice versa:
$$
\frac{a^{-6}}{8 b^{-6}} = \frac{b^{6}}{8 a^{6}}
$$
Now, the expression has only positive exponents.
3Step 3: Simplify the expression
The expression is already simplified and written with only positive exponents:
$$
\frac{b^{6}}{8 a^{6}}
$$
Key Concepts
Positive ExponentsAlgebraic ManipulationExponentiation Rules
Positive Exponents
In mathematical expressions, exponents play a crucial role in describing how many times a number or a variable is used as a factor in a product. Positive exponents are straightforward, indicating that the base is multiplied by itself a number of times equal to the exponent value. For example, the expression \( a^{3} \) means \( a \times a \times a \).
When simplifying expressions, it's essential to work with positive exponents because they are easier to understand and work with algebraically. Whenever you find negative exponents, the goal is to convert them into positive exponents. This is done by applying the rule that tells us \( a^{-n} = \frac{1}{a^{n}} \) if \( n \) is positive. Thus, by moving terms with negative exponents across the fraction bar, the negative sign is eliminated, and we obtain a positive exponent.
When simplifying expressions, it's essential to work with positive exponents because they are easier to understand and work with algebraically. Whenever you find negative exponents, the goal is to convert them into positive exponents. This is done by applying the rule that tells us \( a^{-n} = \frac{1}{a^{n}} \) if \( n \) is positive. Thus, by moving terms with negative exponents across the fraction bar, the negative sign is eliminated, and we obtain a positive exponent.
Algebraic Manipulation
Algebraic manipulation refers to the suite of techniques used to transform equations and expressions into simpler or more useful forms. These skills are particularly essential when dealing with exponents. During algebraic manipulation, you may apply various exponentiation rules, distribute powers over terms, factor expressions, and move terms across the fraction bar to flip the sign of the exponent.
For instance, when given \( \left(\frac{a^{-2}}{2 b^{2}}\right)^{-3} \), you need to use algebraic manipulation to simplify it. In this case, you first apply the power of -3 to each term, which involves using exponentiation rules. Then, to ensure all your exponents are positive, you manipulate the expression by flipping the terms with negative exponents. The simplified form retains the same value as the original expression but is much easier to handle in subsequent calculations.
For instance, when given \( \left(\frac{a^{-2}}{2 b^{2}}\right)^{-3} \), you need to use algebraic manipulation to simplify it. In this case, you first apply the power of -3 to each term, which involves using exponentiation rules. Then, to ensure all your exponents are positive, you manipulate the expression by flipping the terms with negative exponents. The simplified form retains the same value as the original expression but is much easier to handle in subsequent calculations.
Exponentiation Rules
Exponentiation rules are the backbone of operating with powers in algebra. These rules help us to simplify expressions with exponents efficiently. Some fundamental rules include:
- The product rule: \( a^{m} \times a^{n} = a^{(m+n)} \)
- The quotient rule: \( a^{m} ÷ a^{n} = a^{(m-n)} \)
- The power rule: \( (a^{m})^{n} = a^{(m \times n)} \)
- Power of a product: \( (ab)^{n} = a^{n}b^{n} \)
- Power of a quotient: \( \left(\frac{a}{b}\right)^{n} = \frac{a^{n}}{b^{n}} \)
Other exercises in this chapter
Problem 34
Solve the equation by using the quadratic formula. $$ 15-2 y^{2}=7 y $$
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Carry out the indicated operation and write your answer using positive exponents only. $$ \left(\frac{27 x^{-3} y^{2}}{8 x^{-2} y^{-5}}\right)^{1 / 3} $$
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Solve the given equation. $$ \sqrt{2 x-3}-3=0 $$
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In Exercises, factor the polynomial. If the polynomial is prime, state it. $$ 16 u^{4} v-9 v^{3} $$
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