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Chapter 1: Problem 32

Exponents Simplify each expression, and eliminate any negative exponents. (a) \(y^{2} \cdot y^{-5}\) (b) \(z^{5} z^{-3} z^{-4}\) (c) \(\frac{y^{7} y^{0}}{y^{10}}\)

Short Answer

Expert verified
(a) \(\frac{1}{y^{3}}\), (b) \(\frac{1}{z^{2}}\), (c) \(\frac{1}{y^{3}}\).

Step by step solution

01

Simplify Expression (a)

The expression is given by \(y^{2} \cdot y^{-5}\). According to the rules of exponents, when multiplying expressions with the same base, you add the exponents. Thus, \(y^{2} \cdot y^{-5} = y^{2 + (-5)} = y^{-3}\). Finally, to eliminate the negative exponent, rewrite it as \(\frac{1}{y^{3}}\).
02

Simplify Expression (b)

The expression is given by \(z^{5} z^{-3} z^{-4}\). We apply the exponent rule for multiplication again: \(z^{5} \cdot z^{-3} \cdot z^{-4} = z^{5 + (-3) + (-4)} = z^{-2}\). Eliminate the negative exponent: \(z^{-2} = \frac{1}{z^{2}}\).
03

Simplify Expression (c)

The expression is \(\frac{y^{7} y^{0}}{y^{10}}\). First, note that \(y^{0} = 1\) because any number to the power of zero is 1. This simplifies the expression to \(\frac{y^{7} \cdot 1}{y^{10}} = \frac{y^{7}}{y^{10}}\). Using the rule for division of the same base, subtract the exponents: \(y^{7 - 10} = y^{-3}\). Convert the negative exponent to a fraction: \(\frac{1}{y^{3}}\).

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Key Concepts

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

Understanding Negative Exponents
Negative exponents can seem a bit intimidating at first, but they鈥檙e quite simple once you get the hang of them. Essentially, a negative exponent means you need to take the reciprocal of what you have. So, if you see an expression like \(a^{-n}\), that鈥檚 the same as saying \(\frac{1}{a^n}\). This is a very useful transformation to understand because it allows us to rewrite expressions without negative exponents, making them easier to work with and understand.
  • For example, if we have \(y^{-3}\), this is equivalent to \(\frac{1}{y^{3}}\).
  • Similarly, \(z^{-2}\) becomes \(\frac{1}{z^{2}}\).
Remember, when you encounter negative exponents, just flip the base to the denominator with the positive exponent, and you'll be on the right track!
Simplifying Expressions with Exponents
Simplifying expressions with exponents involves combining like terms and using the rules of exponents to make expressions as simple as possible.
Let鈥檚 look at an example to understand this better. If you see something like \(y^{2} \cdot y^{-5}\), you might feel a little lost at first. But remember, when you multiply terms that have the same base, you simply add the exponents together.
  • For \(y^{2} \cdot y^{-5}\), the exponents are \(2\) and \(-5\), added together to get \(2 + (-5) = -3\).
  • This gives us \(y^{-3}\), which we then rewrite as \(\frac{1}{y^{3}}\) by using the rule we just talked about for negative exponents.
By following these steps, you'll be able to simplify parts of expressions easily and rewrite them in their most straightforward form.
The Rules of Exponents
Mastering the rules of exponents is crucial for solving problems involving expressions with powers. Here are some key rules you should always keep in mind when working with exponents:
  • Multiplication of Same Base: When multiplying expressions with the same base, add the exponents: \(a^{m} \cdot a^{n} = a^{m+n}\). This is what we used earlier with \(y^{2} \cdot y^{-5}\).
  • Division of Same Base: For dividing expressions with the same base, subtract the exponents: \(\frac{a^m}{a^n} = a^{m-n}\). This was applied in \(\frac{y^{7}}{y^{10}} = y^{7-10} = y^{-3}\).
  • Zero Exponent Rule: Any base with an exponent of zero is always equal to one, \(a^0 = 1\), which simplifies expressions such as \(y^{7} y^{0}\) by turning \(y^{0}\) into \(1\).
Each of these rules helps transform and simplify expressions, making them easier to handle and understand. With practice, these rules become second nature and help solve even more complex expressions efficiently.

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Most popular questions from this chapter

Solving an Equation in Different Ways We have learned several different ways to solve an equation in this section. Some equations can be tackled by more than one method. For example, the equation \(x-\sqrt{x}-2=0\) is of quadratic type. We can solve it by letting \(\sqrt{x}=u\) and \(x=u^{2},\) and factoring. Or we could solve for \(\sqrt{x},\) square each side, and then solve the resulting quadratic equation. Solve the following equations using both methods indicated, and show that you get the same final answers. (a) \(x-\sqrt{x}-2=0 \quad\) quadratic type; solve for the radical, and square (b) \(\frac{12}{(x-3)^{2}}+\frac{10}{x-3}+1=0 \quad \begin{array}{l}\text { quadratic type; multiply } \\ \text { by } \mathrm{LCD}\end{array}\)

It follows from Kepler's Third Law of planetary motion that the average distance from a planet to the sun (in meters) is $$d=\left(\frac{G M}{4 \pi^{2}}\right)^{1 / 3} T^{2 / 3}$$ where \(M=1.99 \times 10^{30} \mathrm{kg}\) is the mass of the sun, \(G=6.67 \times 10^{-11} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}\) is the gravitational constant, and \(T\) is the period of the planet's orbit (in seconds). Use the fact that the period of the earth's orbit is about 365.25 days to find the distance from the earth to the sun.

Depth of a Well One method for determining the depth of a well is to drop a stone into it and then measure the time it takes until the splash is heard. If \(d\) is the depth of the well (in feet) and \(t_{1}\) the time (in seconds) it takes for the stone to fall, then \(d=16 t_{1}^{2},\) so \(t_{1}=\sqrt{d} / 4 .\) Now if \(t_{2}\) is the time it takes for the sound to travel back up, then \(d=1090 t_{2}\) because the speed of sound is 1090 ft's. So \(t_{2}=d / 1090\) Thus the total time elapsed between dropping the stone and hearing the splash is $$t_{1}+t_{2}=\frac{\sqrt{d}}{4}+\frac{d}{1090}$$ How deep is the well if this total time is 3 s? PICTURE CANT COPY

Radicals Simplify the expression, and eliminate any negative exponents(s). Assume that all letters denote positive numbers. (a) \(\sqrt[3]{y \sqrt{y}}\) (b) \(\sqrt{\frac{16 u^{3} v}{u v^{5}}}\)

A determined gardener has \(120 \mathrm{ft}\) of deer-resistant fence. She wants to enclose a rectangular vegetable garden in her backyard, and she wants the area that is enclosed to be at least \(800 \mathrm{ft}^{2}\). What range of values is possible for the length of her garden?
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