Section 6.2
Integer Exponents
Definition: For $a\ne 0,\ \ {{a}^{0}}=1.$
The rationale behind this definition is that in order to maintain the rule $\frac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$ when $m=n$, we need to have ${{a}^{0}}=1$, since $\frac{{{a}^{m}}}{{{a}^{m}}}=1$ if $a\ne 0$.
Definition: For$a\ne 0,\ \ {{a}^{-n}}=\frac{1}{{{a}^{n}}}$.
Again, the rationale is to preserve our exponent rules. On the one hand, according to the exponent rules, $\frac{{{a}^{0}}}{{{a}^{n}}}={{a}^{0-n}}={{a}^{-n}}$. At the same time, according to our previous definition: $\frac{{{a}^{0}}}{{{a}^{n}}}=\frac{1}{{{a}^{n}}}$. Thus it makes sense to define ${{a}^{-n}}\text{ as }\frac{1}{{{a}^{n}}}$.
Examples:
$\text{a.}\;\;\;{{7}^{-2}}=\dfrac{1}{{{7}^{2}}}=\dfrac{1}{49}$
$\text{b.}\;\;\;\dfrac{1}{{{7}^{-2}}}={{7}^{2}}=49$
$\text{c.}\;\;\;a\cdot {{b}^{-3}}=\dfrac{a}{{{b}^{3}}}$
$\text{d.}\;\;\;\dfrac{{{a}^{-2}}}{{{b}^{-3}}}=\dfrac{{{b}^{3}}}{{{a}^{2}}}$
Note that negative exponents in the numerator result in positive exponents in the denominator; and negative exponents in the denominator produce positive exponents in the numerator. Negative exponents do NOT, as a matter of course, produce negative numbers! That only happens if the quantity being raised to a power (the base) has a negative sign and the exponent is odd (see example g.). There is still no general exponent rule when addition or subtraction occurs before the exponent (see h.).
\begin{align*}
\text{e. }\qquad\dfrac{2{{x}^{10}}{{y}^{2}}}{6{{x}^{5}}{{y}^{5}}}&=\dfrac{{{x}^{5}}{{y}^{-3}}}{3}\\
&=\dfrac{{{x}^{5}}}{3{{y}^{3}}}
\end{align*}
\begin{align*}
\text{f. }\qquad\dfrac{{{\left( 2{{x}^{-2}}y \right)}^{3}}}{4{{x}^{2}}y}&=\dfrac{{{2}^{3}}{{x}^{-6}}{{y}^{3}}}{{{2}^{2}}{{x}^{2}}y}\\
&=\dfrac{2{{y}^{2}}}{{{x}^{8}}}
\end{align*}
\begin{align}
\text{g. }\qquad{{\left( -2{{x}^{-3}} \right)}^{-3}}&={{(-2)}^{-3}}\cdot {{\left( {{x}^{-3}} \right)}^{-3}} \\
& =\frac{1}{{{(-2)}^{3}}}\cdot {{x}^{9}}\\
&=-\frac{{{x}^{9}}}{8}
\end{align}
\begin{align}
\text{h. }\qquad{{\left( {{2}^{-1}}-{{3}^{-1}} \right)}^{-1}}&={{\left( \frac{1}{2}-\frac{1}{3} \right)}^{-1}} \\
& ={{\left( \frac{1}{6} \right)}^{-1}}\\
&=6
\end{align}
Negative exponents:
Exercises:
Simplify each of the following. Variables represent positive numbers. Write all answers with positive exponents.
\begin{array}{l l l}
1. \;\;\;{{\left( 2{{x}^{-2}} \right)}^{-1}}\;\;\; & 2. \;\;\;{{\left( -3{{x}^{2}}{{y}^{-3}} \right)}^{-2}}\;\;\;& 3. \;\;\;2{{x}^{2}}{{y}^{-3}}\cdot {{\left( 3{{x}^{2}}y \right)}^{2}}\;\;\;\\
4. \;\;\;\left( 2{{x}^{2}}{{y}^{3}} \right)\cdot {{\left( 2{{x}^{-2}}{{y}^{-3}} \right)}^{2}}\;\;\; & 5. \;\;\;{{\left( 2{{x}^{-2}}{{y}^{3}} \right)}^{-2}}\;\;\; & 6. \;\;\;{{\left( {{x}^{-2}}\cdot {{x}^{-1}} \right)}^{-2}}\cdot {{\left( x\cdot {{x}^{2}}\cdot {{x}^{3}} \right)}^{2}}\;\;\; \\
7. \;\;\;{{\left( 2{{x}^{2}}{{y}^{-3}} \right)}^{4}}\;\;\; & 8. \;\;\;{{\left( 2{{x}^{2}}{{y}^{3}} \right)}^{0}}\cdot {{\left( 3{{x}^{3}}{{y}^{-2}} \right)}^{-2}}\;\;\; & 9. \;\;\;{{\left( 2{{x}^{2}}{{y}^{-3}} \right)}^{-2}}\;\;\; \\
10. \;\;\;{{\left( 2{{x}^{2}}y \right)}^{-3}}\cdot {{\left( 4{{x}^{3}}y \right)}^{2}}\;\;\; & 11. \;\;\;\frac{1}{3{{x}^{-2}}}\;\;\; & 12. \;\;\;\frac{{{x}^{-3}}y}{6{{y}^{-3}}}\;\;\; \\
13. \;\;\;\frac{{{\left( 2{{x}^{2}}{{y}^{-3}} \right)}^{2}}}{{{\left( 6{{x}^{-2}}\cdot {{y}^{3}} \right)}^{-2}}}\;\;\; & 14. \;\;\;{{\left( \frac{2{{x}^{-3}}{{y}^{2}}}{{{x}^{2}}y} \right)}^{4}}\;\;\; & 15. \;\;\;\frac{1}{x}\cdot {{\left( \frac{{{x}^{2}}}{2} \right)}^{-1}}\cdot {{\left( \frac{3}{{{x}^{3}}} \right)}^{-2}}\;\;\; \\
16. \;\;\;{{\left( \frac{2{{x}^{3}}{{y}^{-3}}}{6{{x}^{-2}}y} \right)}^{-1}}\;\;\; & 17. \;\;\;{{\left( \frac{12{{x}^{-2}}{{y}^{3}}}{{{2}^{-1}}{{x}^{2}}y} \right)}^{-1}}\;\;\; & 18. \;\;\;{{\left( \frac{4{{x}^{2}}{{y}^{-3}}}{{{2}^{-1}}{{x}^{-2}}{{y}^{-2}}} \right)}^{-2}}\;\;\; \\
19. \;\;\;{{\left( \frac{{{4}^{-2}}{{x}^{2}}}{2{{y}^{-2}}} \right)}^{-1}}\cdot \left( \frac{{{x}^{2}}{{y}^{2}}}{8} \right)\;\;\; & 20. \;\;\;{{\left( \frac{3}{{{9}^{-2}}{{K}^{2}}} \right)}^{-3}}\cdot \left( \frac{{{K}^{2}}}{27} \right)\;\;\; & 21. \;\;\;{{\left( \frac{{{3}^{-2}}{{a}^{-1}}}{9{{K}^{2}}} \right)}^{-2}}\cdot \left( \frac{{{a}^{-2}}{{K}^{2}}}{27} \right)\;\;\; \\
22. \;\;\;{{x}^{-3}}\cdot {{x}^{m-1}}\cdot {{x}^{4+m}}\;\;\; & 23. \;\;\;{{x}^{-3}}\cdot {{x}^{m-1}}\div \left( {{x}^{4+m}}\cdot {{x}^{5}} \right)\;\;\; & 24. \;\;\;\frac{{{x}^{-2}}{{y}^{-3}}}{{{\left( {{x}^{2}}y \right)}^{-1}}}\cdot {{\left( \frac{y}{x} \right)}^{-3}}\;\;\; \\
25. \;\;\;\frac{{{({{2}^{-2}}{{x}^{2}}{{y}^{-3}})}^{-1}}}{4{{x}^{2}}{{y}^{-1}}}\;\;\; & 26. \;\;\;{{x}^{-3}}\cdot {{x}^{m-1}}\div {{x}^{4+m}}\cdot {{x}^{5}}\;\;\; & 27. \;\;\;\frac{{{({{2}^{-2}}{{x}^{2}}{{y}^{-3}})}^{2}}}{{{\left( 4{{x}^{2}}{{y}^{-1}} \right)}^{-2}}}\;\;\; \\
28. \;\;\;{{(a-b)}^{-1}}\;\;\; & 29. \;\;\;\frac{{{x}^{2m+1}}\cdot {{x}^{2-m}}}{{{x}^{3}}}\;\;\; & 30. \;\;\;\frac{2{{x}^{2}}\cdot {{\left( 2{{x}^{2}} \right)}^{-1}}\cdot 2{{\left( {{x}^{2}} \right)}^{2}}}{4{{x}^{2}}\cdot {{\left( 4{{x}^{2}} \right)}^{0}}}\;\;\; \\
31. \;\;\;{{({{a}^{-1}}+b)}^{-1}}\;\;\; & 32. \;\;\;\frac{{{({{a}^{-1}}+b)}^{-1}}}{ab}\;\;\; & 33. \;\;\;{{(x+1)}^{-1}}\cdot {{(x+1)}^{0}}\cdot {{\left( x+1 \right)}^{2}}\;\;\; \\
34. \;\;\;\frac{{{2}^{-1}}{{x}^{-3}}{{y}^{2}}}{{{2}^{-2}}{{x}^{-3}}{{y}^{-1}}}\;\;\; & 35. \;\;\;{{\left( \frac{{{m}^{x-1}}\cdot {{m}^{x-2}}}{{{m}^{2x-5}}} \right)}^{-3}}\;\;\; & 36. \;\;\;\left( {{2}^{-1}}-{{3}^{-1}} \right)\div \left( {{4}^{-1}}-{{5}^{-1}} \right)\;\;\; \\
37. \;\;\;{{(2a-b)}^{-2}}\;\;\; & 38. \;\;\;{{\left( {{a}^{-1}}\div {{b}^{-2}} \right)}^{-1}}\;\;\; & 39. \;\;\;{{(x+1)}^{-1}}\cdot {{(x+1)}^{0}}\cdot \left( {{x}^{-1}}+1 \right)\;\;\; \\
40. \;\;\;{{\left( {{a}^{-1}}-{{b}^{-2}} \right)}^{-1}}\;\;\; & 41. \;\;\;{{\left( {{a}^{-1}}\cdot {{b}^{-2}} \right)}^{-1}}\;\;\; & 42. \;\;\;{{(-2)}^{-1}}-\frac{1}{{{\left( {{x}^{-1}}-{{2}^{-1}} \right)}^{-1}}}\;\;\; \\
\end{array}
Write each of the following expressions with exponents in such a way that there are no fractions. With the numbers, use powers of primes only (that is, write ${{2}^{2}}\cdot 3$ rather than $12$).
\begin{array}{l l l}
43. \;\;\;\frac{{{x}^{3}}{{y}^{5}}}{2x{{y}^{7}}}\;\;\; & 44. \;\;\;\frac{2}{8{{x}^{5}}{{y}^{-6}}}\;\;\; & 45. \;\;\;\frac{6}{{{\left( 2{{x}^{-2}}{{y}^{3}} \right)}^{-2}}}\;\;\; \\
46. \;\;\;\frac{10{{x}^{13}}{{y}^{5}}}{25{{x}^{5}}{{y}^{6}}}\;\;\; & 47. \;\;\;\frac{18{{x}^{3}}{{y}^{5}}}{4{{x}^{5}}{{y}^{17}}}\;\;\; & 48. \;\;\;\frac{{{\left( {{4}^{2}}{{x}^{-2}}{{y}^{5}} \right)}^{3}}}{8{{x}^{5}}{{y}^{-6}}}\;\;\;
\end{array}
49. Identify each of the following as TRUE or FALSE.
\begin{array}{l l l}
a. \;\;\;2{{x}^{0}}=2\;\;\; & b. \;\;\;2{{x}^{-1}}=\frac{1}{2x}\;\;\; & c. \;\;\;{{\left( {{x}^{2}}+y \right)}^{2}}={{x}^{4}}+{{y}^{2}}\;\;\; \\
d. \;\;\;{{\left( {{2}^{3}} \right)}^{4}}={{\left( {{2}^{4}} \right)}^{3}}\;\;\; & e. \;\;\;{{2}^{2}}\cdot {{2}^{3}}\cdot {{2}^{4}}\cdot {{2}^{5}}={{16}^{14}}\;\;\; & f. \;\;\;{{\left( {{x}^{-1}}+{{y}^{-1}} \right)}^{-1}}=x+y\;\;\; \\
g. \;\;\;\frac{y}{{{x}^{-1}}}=xy\;\;\; & h. \;\;\;{{\left( -2{{x}^{-2}} \right)}^{-2}}=\frac{1}{4}{{x}^{4}}\;\;\; & i. \;\;\;\frac{{{x}^{-3}}}{{{x}^{-5}}}={{x}^{2}}\;\;\; \\
j. \;\;\;{{\left( {{x}^{3}} \right)}^{4}}={{x}^{7}}\;\;\; & k. \;\;\;2{{x}^{-4}}=\frac{1}{16{{x}^{4}}}\;\;\; & l. \;\;\;(x-1)\cdot x\cdot (x+1)={{x}^{3}}-1\;\;\;
\end{array}
50. The only prime factors of 2, 4, 6, 7, 9, and 12 are 2 and 3. If ${{2}^{20}}\cdot {{4}^{10}}\cdot {{6}^{5}}\cdot {{8}^{3}}\cdot {{9}^{2}}\cdot {{12}^{12}}={{2}^{x}}\cdot {{3}^{y}}$, then what are the values of $x$ and $y$?