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Use the exponent rules to simplify the following expressions: \(\left(-\frac{x^{-4} y^{-4}}{x^{2} y^{-2}}\right)^{5}\)

Short Answer

Expert verified
The simplified expression is \(-\frac{y^{10}}{x^{30}}\)."

Step by step solution

01

Rewrite the Expression Using Positive Exponents

The expression given is \( \left(-\frac{x^{-4} y^{-4}}{x^{2} y^{-2}}\right)^{5} \). Rewriting the exponents as positive exponents, we get: \(-\frac{1}{x^{4} y^{4}} \cdot \frac{1}{x^2 y^{-2}} = -\frac{y^2}{x^{6}} \).
02

Apply the Power of a Power Rule

Apply the power of a power rule \((a^m)^n = a^{m \cdot n}\) on each part inside the parentheses: \(\left(-\frac{y^2}{x^6} \right)^{5} = -\frac{y^{2\cdot 5}}{x^{6\cdot 5}} = -\frac{y^{10}}{x^{30}}\).
03

Simplify the Expression

The expression \(-\frac{y^{10}}{x^{30}}\) is already simplified, as both exponents are positive and there are no common terms to reduce further.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Simplifying Expressions
Simplifying expressions is a process of rewriting them in a form that is easier to work with or understand. This often involves using the rules of arithmetic operations and managing expressions to make calculations more straightforward.
When simplifying expressions that include exponents, like \( \left(-\frac{x^{-4}y^{-4}}{x^{2}y^{-2}}\right)^{5} \), we need to pay attention to key rules that make the handling of such expressions easy. These expressions often have negative exponents, fractions, or are expressed as powers of powers.
  • Start by converting any negative exponents into positive ones. This means flipping the terms; those with positive exponents go to the numerator, and those with negative exponents go to the denominator.
  • Follow with simplification by factoring and reducing where possible.
Mastering the art of simplification is not only about getting the right answer but doing it in a way that reveals the most about the mathematical relationships involved.
Positive Exponents
When dealing with exponents, it is often necessary to convert negative exponents into positive ones to simplify calculations.
The rule to remember is: \(a^{-n} = \frac{1}{a^n}\). By applying this rule, you turn terms like \(x^{-4}\) into \(\frac{1}{x^4}\). This makes managing complex algebraic expressions much simpler.
  • Always aim to convert expressions with negative exponents first, as they tend to complicate equations.
  • In fractional expressions, rewrite the numbers so that all the exponents are positive by reversing their placement between numerator and denominator.
Consider in the problem given: the initial step transforms \(\left(-\frac{x^{-4} y^{-4}}{x^{2} y^{-2}}\right)\)into\(-\frac{1}{x^4 y^4} \cdot y^2\). Understanding this step is crucial to solving the expression efficiently.
Power of a Power Rule
The Power of a Power Rule is an important rule when simplifying complex expressions. Its formula is \((a^m)^n = a^{m \cdot n}\). This means that when you have an exponent raised by another exponent, you multiply the exponents together.
In our exercise, after converting all exponents to positive, the Power of a Power Rule helps in reducing the expression\(\left(-\frac{y^2}{x^6} \right)^5\)into \(-\frac{y^{2\cdot5}}{x^{6\cdot5}} = -\frac{y^{10}}{x^{30}}\).
  • Applying this rule is simple: identify the "base" and "exponents", then multiply them.
  • This rule is particularly useful to avoid expansive multiplication, streamlining our calculations.
Understanding and applying this rule allows for a quick reduction in complex expressions, preserving significant relationships within the algebraic terms.

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Most popular questions from this chapter

If an object is moving at a constant speed \(v,\) then the time the object takes to traverse a distance \(d\) is inversely proportional to its speed \(v .\) If it takes a car \(1 \mathrm{~h}\) to travel a distance of \(60 \mathrm{mi},\) how long will it take the car to travel the same distance if it slows down to one-third of its original speed?

For a sound coming from a point source, the amplitude of sound is inversely proportional to the distance. If the displacement amplitude of an air molecule in a sound wave is \(4.8 \times 10^{-6} \mathrm{~m}\) at a point \(1.0 \mathrm{~m}\) from the source, what would be the displacement amplitude of the same sound when the distance increases to \(4.0 \mathrm{~m} ?\)

According to the ideal-gas law (Section 15.2 ), the volume of an ideal gas is directly proportional to its temperature in kelvins (K) if the pressure of the gas is constant. An ideal gas occupies a volume of 4.0 liters at \(100 \mathrm{~K}\). Determine its volume when it is heated to \(300 \mathrm{~K}\) while held at a constant pressure.

The force of gravity on an object (which we experience as the object's weight) varies inversely as the square of the distance from the center of the earth. Determine the force of gravity on an astronaut when he is at a height of \(6000 \mathrm{~km}\) from the surface of the earth if he weighs 700 newtons (N) when on the surface of the earth. The radius of the earth is \(6.38 \times 10^{6} \mathrm{~m}\). (If the astronaut is in orbit, he will float "weightlessly," but gravity still acts on him- he and his spaceship appear weightless because they are falling freely in their orbit around the earth.)

Solve the following equations using any method: \(5 x-4 y=1,6 y=10 x-4\)

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