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Обучение Руководства > College Algebra

Using the Negative Rule of Exponents

Another useful result occurs if we relax the condition that m>nm>n in the quotient rule even further. For example, can we simplify h3h5\frac{{h}^{3}}{{h}^{5}}? When m<nm<n—that is, where the difference mnm-n is negative—we can use the negative rule of exponents to simplify the expression to its reciprocal. Divide one exponential expression by another with a larger exponent. Use our example, h3h5\frac{{h}^{3}}{{h}^{5}}.
h3h5=hhhhhhhh=hhhhhhhh=1hh=1h2\begin{array}{ccc}\hfill \frac{{h}^{3}}{{h}^{5}}& =& \frac{h\cdot h\cdot h}{h\cdot h\cdot h\cdot h\cdot h}\hfill \\ & =& \frac{\cancel{h}\cdot \cancel{h}\cdot \cancel{h}}{\cancel{h}\cdot \cancel{h}\cdot \cancel{h}\cdot h\cdot h}\hfill \\ & =& \frac{1}{h\cdot h}\hfill \\ & =& \frac{1}{{h}^{2}}\hfill \end{array}
If we were to simplify the original expression using the quotient rule, we would have
h3h5=h35= h2\begin{array}{ccc}\hfill \frac{{h}^{3}}{{h}^{5}}& =& {h}^{3 - 5}\hfill \\ & =& \text{ }{h}^{-2}\hfill \end{array}
Putting the answers together, we have h2=1h2{h}^{-2}=\frac{1}{{h}^{2}}. This is true for any nonzero real number, or any variable representing a nonzero real number. A factor with a negative exponent becomes the same factor with a positive exponent if it is moved across the fraction bar—from numerator to denominator or vice versa.
an=1anandan=1an\begin{array}{ccc}{a}^{-n}=\frac{1}{{a}^{n}}& \text{and}& {a}^{n}=\frac{1}{{a}^{-n}}\end{array}
We have shown that the exponential expression an{a}^{n} is defined when nn is a natural number, 0, or the negative of a natural number. That means that an{a}^{n} is defined for any integer nn. Also, the product and quotient rules and all of the rules we will look at soon hold for any integer nn.

A General Note: The Negative Rule of Exponents

For any nonzero real number aa and natural number nn, the negative rule of exponents states that
an=1an{a}^{-n}=\frac{1}{{a}^{n}}

Example 5: Using the Negative Exponent Rule

Write each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents.
  1. θ3θ10\frac{{\theta }^{3}}{{\theta }^{10}}
  2. z2zz4\frac{{z}^{2}\cdot z}{{z}^{4}}
  3. (5t3)4(5t3)8\frac{{\left(-5{t}^{3}\right)}^{4}}{{\left(-5{t}^{3}\right)}^{8}}

Solution

  1. θ3θ10=θ310=θ7=1θ7\frac{{\theta }^{3}}{{\theta }^{10}}={\theta }^{3 - 10}={\theta }^{-7}=\frac{1}{{\theta }^{7}}
  2. z2zz4=z2+1z4=z3z4=z34=z1=1z\frac{{z}^{2}\cdot z}{{z}^{4}}=\frac{{z}^{2+1}}{{z}^{4}}=\frac{{z}^{3}}{{z}^{4}}={z}^{3 - 4}={z}^{-1}=\frac{1}{z}
  3. (5t3)4(5t3)8=(5t3)48=(5t3)4=1(5t3)4\frac{{\left(-5{t}^{3}\right)}^{4}}{{\left(-5{t}^{3}\right)}^{8}}={\left(-5{t}^{3}\right)}^{4 - 8}={\left(-5{t}^{3}\right)}^{-4}=\frac{1}{{\left(-5{t}^{3}\right)}^{4}}

Try It 5

Write each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents.

a. (3t)2(3t)8\frac{{\left(-3t\right)}^{2}}{{\left(-3t\right)}^{8}} b. f47f49f\frac{{f}^{47}}{{f}^{49}\cdot f} c. 2k45k7\frac{2{k}^{4}}{5{k}^{7}}

Solution

Example 6: Using the Product and Quotient Rules

Write each of the following products with a single base. Do not simplify further. Write answers with positive exponents.
  1. b2b8{b}^{2}\cdot {b}^{-8}
  2. (x)5(x)5{\left(-x\right)}^{5}\cdot {\left(-x\right)}^{-5}
  3. 7z(7z)5\frac{-7z}{{\left(-7z\right)}^{5}}

Solution

  1. b2b8=b28=b6=1b6{b}^{2}\cdot {b}^{-8}={b}^{2 - 8}={b}^{-6}=\frac{1}{{b}^{6}}
  2. (x)5(x)5=(x)55=(x)0=1{\left(-x\right)}^{5}\cdot {\left(-x\right)}^{-5}={\left(-x\right)}^{5 - 5}={\left(-x\right)}^{0}=1
  3. 7z(7z)5=(7z)1(7z)5=(7z)15=(7z)4=1(7z)4\frac{-7z}{{\left(-7z\right)}^{5}}=\frac{{\left(-7z\right)}^{1}}{{\left(-7z\right)}^{5}}={\left(-7z\right)}^{1 - 5}={\left(-7z\right)}^{-4}=\frac{1}{{\left(-7z\right)}^{4}}

Try It 6

Write each of the following products with a single base. Do not simplify further. Write answers with positive exponents.
  1. t11t6{t}^{-11}\cdot {t}^{6}
  2. 25122513\frac{{25}^{12}}{{25}^{13}}
Solution

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