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Chapter 7: Problem 13

Simplify each of the following expressions as completely as possible. Final answers should be expressed with positive exponents only. (Assume that all variables represent positive quantities.) $$5^{-2}$$

Short Answer

Expert verified
\[5^{-2} = \frac{1}{25}\]

Step by step solution

01

Understand the negative exponent

A negative exponent indicates that the base should be reciprocated. For example, \(a^{-n} = \frac{1}{a^n}\). In this case, the base is 5 and the exponent is -2.
02

Apply the reciprocal rule

Reciprocate the base 5 to remove the negative exponent. This gives us: \[5^{-2} = \frac{1}{5^2}\]
03

Simplify the expression

Now calculate the value of the denominator: \[5^2 = 25\]. Hence, the expression simplifies to: \[5^{-2} = \frac{1}{25}\]

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Key Concepts

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

Negative Exponents
Negative exponents indicate that the base number should be reciprocated or taken to the reciprocal. When you see a negative exponent, such as in the expression \(5^{-2}\), it means you'll need to 'flip' the base.

For example: \(a^{-n} \) equals \( \frac{1}{a^n} \). In simpler terms, instead of multiplying the base by itself a certain number of times, you divide one by that multiplication.

In our example: \(5^{-2} \), the base is 5, and the exponent is -2. This means we should write it as \( \frac{1}{5^2} \).
Reciprocal Rule
The reciprocal rule transforms expressions with negative exponents into fractions. Once you see a negative exponent, apply this rule to rewrite it.

Take the base, for instance 5, and turn the exponent positive. This flips the base into the denominator of a fraction, leaving 1 as the numerator.

For example: \(5^{-2} \) becomes \( \frac{1}{5^2} \). Now the expression is much easier to manage.

After applying the reciprocal rule, the negative exponent is 'gone,' and you can proceed to simplify the expression with positive exponents.
Positive Exponents
Exponents are used to indicate how many times a number (the base) is multiplied by itself. Positive exponents are straightforward. If you have \(5^2\), it means 5 multiplied by itself 2 times: 5 * 5.

To simplify the expression \( \frac{1}{5^2} \):
  • Calculate \(5^2 \), which gives us 25
  • Thus the expression becomes \( \frac{1}{25} \).

    When simplifying expressions, remember to always convert any negative exponents to their positive counterparts before fully simplifying the expression.

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Most popular questions from this chapter

Perform the indicated operations and simplify. $$\frac{a}{3}-\frac{a}{2}+\frac{a}{5}$$

Explain how you would multiply \(3.74 \times 10^{-5}\) by \(6.38 \times 10^{4}\) without converting either number into standard notation.

Multiply out each of the following. As you work out the problems, identify those exercises that are either a perfect square or the difference of two squares. $$(2 y-5)(4 y-3)$$

Estimate the answer without actually carrying out the computation and make the most appropriate choice. If you multiply 0.00032 by \(420,000,\) the result is closest to (a) 0.12 (b) 1.2 (c) 12 (d) 120 (e) 1200

In this section we defined \(a^{0}=1 .\) It is important to realize that a definition is neither right nor wrong-it just is. The proper question to ask about a definition is "Is it useful?" What happens if someone decides on an "alternative definition" such as \(a^{0}=7\) because 7 happens to be his or her favorite number? What happens when we consider the expression \(a^{0} \cdot a^{4} ?\) If you use exponent rule 1 you get \(a^{0} \cdot a^{4}=a^{0+4}=a^{4} .\) However, if you use this alternative definition, you get \(a^{0} \cdot a^{4} \stackrel{\underline{2}}{=} 7 a^{4} .\) The answer we get from the alternative definition is not consistent with the answer we get from the exponent rule. The exponent rules and this alternative definition cannot coexist. Since we do not want to throw away all the exponent rules, we must modify the alternative definition so that it is consistent with all the exponent rules. Verify that our definition of \(a^{0}=1\) is consistent with all five exponent rules.
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