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Chapter 4: Problem 18

Using rules of exponents, show that \(\frac{9^{5}}{27^{-7}}=3^{31}\).

Short Answer

Expert verified
\(\frac{9^5}{27^{-7}} = 3^{31}\)

Step by step solution

01

Rewrite bases using prime factorization

Rewrite 9 and 27 using their prime base forms. Note that 9 can be written as \(3^2\) and 27 as \(3^3\). Thus, the equation becomes: \(\frac{(3^2)^5}{(3^3)^{-7}}\).
02

Apply power of a power rule

Use the power of a power rule \((a^m)^n = a^{m \times n}\) on both the numerator and the denominator. This gives: \(\frac{3^{2 \times 5}}{3^{3 \times -7}}\).
03

Simplify the exponents

Calculate the exponents: \(2 \times 5 = 10\) and \(3 \times -7 = -21\). Therefore, the equation now is \(\frac{3^{10}}{3^{-21}}\).
04

Apply the quotient rule

Use the quotient rule \(\frac{a^m}{a^n} = a^{m-n}\). Hence: \(\frac{3^{10}}{3^{-21}} = 3^{10 - (-21)}\).
05

Simplify the final exponent

Simplify the exponent subtraction: \(10 - (-21) = 10 + 21 = 31\). Therefore, the final result is \(3^{31}\).

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Key Concepts

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

Prime Factorization
Prime factorization is the process of breaking down a composite number into its prime factors. A prime factor is a prime number that divides the composite number exactly. For example, the number 9 can be written as the product of 3 times 3, which is represented as \(3^2\), and 27 can be written as \(3^3\) because it is the product of 3 multiplied by itself three times. By rewriting numbers using their prime factors, especially in problems involving exponents, it becomes easier to simplify or manipulate the expressions.
Power of a Power Rule
The power of a power rule is an essential tool when working with exponents. This rule states that when you raise a power to another power, you simply multiply the exponents. In mathematical terms, \[(a^m)^n = a^{m \times n}\]. For instance, in the given exercise, we used this rule to transform \(\frac{(3^2)^5}{(3^3)^{-7}}\). By applying the power of a power rule, this becomes \(\frac{3^{2 \times 5}}{3^{3 \times -7}}\), which simplifies to \(\frac{3^{10}}{3^{-21}}\). The rule helps condense and simplify expressions by reducing multiple layers of exponents into a single term.
Quotient Rule
The quotient rule is a fundamental principle for dividing expressions with the same base raised to exponents. The rule states that \(\frac{a^m}{a^n} = a^{m-n}\). When you divide two exponents with the same base, you subtract the exponent in the denominator from the exponent in the numerator. In our exercise, we had \(\frac{3^{10}}{3^{-21}}\). Using the quotient rule simplifies this to \(3^{10 - (-21)}\), which further simplifies to \(3^{31}\) when you perform the subtraction \(10 - (-21)\). This rule is particularly useful for simplifying complex fractional exponent expressions.
Simplifying Exponents
Simplifying exponents involves using different exponent rules to make an expression as straightforward as possible. It may incorporate several steps such as using prime factorization, applying the power of a power rule, and utilizing the quotient rule. In the provided exercise, each step progressively simplified the expression:
  • Rewriting the bases as prime factors: \(\frac{(3^2)^5}{(3^3)^{-7}}\)
  • Applying the power of a power rule: \(\frac{3^{10}}{3^{-21}}\)
  • Employing the quotient rule: \(3^{10 - (-21)}\)
  • Simplifying the resulting exponent: \(3^{31}\)
These techniques are interconnected, and mastery of them enables solving various exponent-related problems efficiently.

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Most popular questions from this chapter

(Requires a calculator that can evaluate powers.) Calculators and spreadsheets use slightly different formats for scientific notation. For example, if you type in Avogadro's number either as 602,000,000,000,000,000,000,000 or as \(6.02 \cdot 10^{23},\) the calculator or spreadsheet will display \(6.02 \mathrm{E} 23,\) where \(\mathrm{E}\) stands for "exponent" or power of \(10 .\) Perform the following calculations using technology, then write the answer in standard scientific notation rounded to three places. a. \(\left(\frac{9}{5}\right)^{50}\) b. \(2^{35}\) c. \(\left(\frac{1}{3}\right)^{7}\) d. \(\frac{7}{6^{15}}\) e. \((5)^{-10}(2)^{10}\) f. \((-4)^{5}\left(\frac{1}{(16)^{12}}\right)\)

Each of the following simplifications contains an error made by students on a test. Find the error and correct the simplification so that the expression becomes true. a. \(\left[\left(x^{2}\right)^{3}\right]^{5}=\left[x^{5}\right]^{5}=x^{25}\) b. \(\frac{7 x^{2} y^{6}}{(x y)^{2}}=\frac{7 x^{2} y^{6}}{x^{2} y^{2}}=7 x^{4} y^{8}\) c. \(\left(\frac{4 x^{3} y^{5}}{6 x y^{4}}\right)^{3}=\left(\frac{2 x^{2} y}{3}\right)^{3}=\frac{2}{3} x^{6} y^{3}\) d. \(\left(1.1 \cdot 10^{6}\right) \cdot\left(1.1 \cdot 10^{4}\right)=1.1 \cdot 10^{6}\) e. \(\frac{4 \cdot 10^{6}}{8 \cdot 10^{3}}=0.5 \cdot 10^{3}=5.0 \cdot 10^{4}\) f. \(6 \cdot 10^{3}+7 \cdot 10^{5}=13 \cdot 10^{8}\)

Solve for \(x .\) (Hint: Rewrite each expression so that you can use a calculator to solve for \(x\).) a. \(\log x=0.82\) b. \(10^{x}=0.012\) c. \(\log x=0.33\) d. \(10^{x}=0.25\)

The distance that light travels in 1 year (a light year) is \(5.88 \cdot 10^{12}\) miles. If a star is \(2.4 \cdot 10^{8}\) light years from Earth, what is this distance in miles?

Change each number to scientific notation, then simplify using rules of exponents. Show your work, recording your final answer in scientific notation. a. \(10 \%\) of 0.00001 b. \(\frac{0.00005}{50,000}\) c. \(\frac{3}{0.006}\) d. \(\frac{8000}{0.0008}\) e. \(\frac{0.0064}{8000}\) f. \(5,000,000 \cdot 40,000\)
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