Chapter 11: Problem 26
Express each of the given expressions in simplest form with only positive exponents. $$5\left(\frac{2 n^{-2}}{D^{-1}}\right)^{-2}$$
Short Answer
Expert verified
The simplest form is \( \frac{5 n^4}{4 D^2} \).
Step by step solution
01
Simplify the Fraction
The expression given is \( 5 \left( \frac{2n^{-2}}{D^{-1}} \right)^{-2} \). We can simplify the fraction \( \frac{2n^{-2}}{D^{-1}} \) as follows. Recognize the negative exponent in \( n^{-2} \) indicates \( \frac{1}{n^{2}} \), so it becomes \( \frac{2}{n^2} \). Since \( D^{-1} \) means \( \frac{1}{D} \), the fraction becomes \( \frac{2n^{-2}}{D^{-1}} = 2n^{-2}D \).
02
Apply the Negative Exponent Outside the Fraction
Now, recall the expression \( 5 \left( 2n^{-2} D \right)^{-2} \). The \(-2\) outside the parentheses means we take the reciprocal and square each term inside the parentheses. Therefore, \( \left( 2n^{-2}D \right)^{-2} = \left( 2^{-1} n^2 D^{-1} \right)^2 = \frac{1}{2^2} n^{2 \times 2} D^{-2} \).
03
Simplify the Expression
Upon expansion, \( \left( 2^{-1} n^2 D^{-1} \right)^2 \) becomes \( \frac{1}{4} n^{4} D^{-2} \). Therefore, \( 5 \left( \frac{2n^{-2}}{D^{-1}} \right)^{-2} \) is equal to \( 5 \times \frac{1}{4} n^{4} D^{-2} \).
04
Multiply the Constant
Finally, multiply the constant 5 by \( \frac{1}{4} \): resulting in \( \frac{5}{4} n^{4} D^{-2} \). To express with positive exponents, rewrite \( D^{-2} \) as \( \frac{1}{D^{2}} \) so the final expression becomes \( \frac{5}{4} n^{4} \times \frac{1}{D^2} \).
05
Final Simplified Expression
Repeat the process to find the expression with positive exponents: The final simplified version is \( \frac{5 n^4}{4 D^2} \), which is in simplest form with only positive exponents.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Negative Exponents
When you see a negative exponent, it might seem confusing at first, but it's quite simple once you break it down. A negative exponent tells you to take the reciprocal of the base. For example, in the expression \( n^{-2} \), the negative exponent \(-2\) means you should write it as \( \frac{1}{n^2} \). This means instead of repeating the base \( n \) in the numerator, it 'flips' it to be in the denominator.
- Negative exponents turn the number upside down. \( a^{-b} = \frac{1}{a^b} \).
- This helps in simplifying expressions and moving terms from numerator to denominator and vice versa.
- It's visually like flipping the term across the fraction bar."
Fraction Simplification
Simplifying fractions ensures that you reduce a fraction to its lowest possible terms. In expressions involving variables and exponents, it involves using algebraic rules. Take the expression \( \frac{2n^{-2}}{D^{-1}} \). By rewriting using exponent rules, here's how it breaks down:
- The term \( n^{-2} \) is rewritten as \( \frac{1}{n^2} \), meaning \( 2n^{-2} \) becomes \( \frac{2}{n^2} \).
- Similarly, \( D^{-1} \) converts into \( \frac{1}{D} \), flipping it when moving to the numerator, yielding \( 2n^{-2}D \).
Positive Exponents
Working with positive exponents keeps things more straightforward. Positive exponents indicate multiplication by the base a certain number of times. For example, in \( n^4 \), you multiply \( n \) by itself four times. It's like a repetitive multiplication, which is easy to understand:
- \( n^4 \) means \( n \times n \times n \times n \).
- Positive numbers stand firm and straightforward, providing clarity in expressions.
- Understanding this makes transitioning from negative to positive exponents simple, you just 'flip' them when simplifying.
Mathematical Expressions
Mathematical expressions encompass a wide range of operations, symbols, and numbers, and they tell a story about numbers and relationships. In the context of this problem, our expression is described first by its terms and operations:
- It involves multiplying, taking powers, and simplifying fractions.
- The expression \( 5\left( \frac{2n^{-2}}{D^{-1}} \right)^{-2} \) shows a chain of operations, like fractions within a power, needing simplification.
- Solving entails breaking each part into understandable pieces before recomposing them into a clearer form.