Exercises: Power Laws

Apply the power laws to simplify the following expressions:

$4^{2} \cdot 4^{9} \cdot 4^{- 12}$

$4^{8} \cdot 2^{- 3} \cdot 2^{5} \cdot 5^{9}$

$\frac{\frac{26^{2}}{26^{8}}}{\frac{13^{- 3}}{13^{3}}}$

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$3^{2} \cdot 3^{1}$	$=$
	↓	Use the power law $a^{x} \cdot a^{y} = a^{x + y}$ with $x = 2, y = 1, a = 3$ .
	$=$	$3^{2 + 1}$
	↓	Conclude the exponent.
	$=$	$3^{3}$

$4^{2} \cdot 4^{9} \cdot 4^{- 12}$	$=$
	↓	First apply the power laws to $4^{2} \cdot 4^{9}$ .
	$=$	$4^{2 + 9} \cdot 4^{- 12}$
	$=$	$4^{11} \cdot 4^{- 12}$
	↓	Now apply the power law to $4^{11} \cdot 4^{- 12}$ .
	$=$	$4^{- 1}$
	↓	The term can be simplified even further by using the rule of the negative exponent, that is $a^{- 1} = \frac{1}{a} .$
	$=$	$\frac{1}{4}$

$4^{8} \cdot 2^{- 3} \cdot 2^{5} \cdot 5^{9}$	$=$
	↓	Apply the power law $a^{x} \cdot a^{y} = a^{x + y}$ to $2^{- 3} \cdot 2^{5}$ .
	$=$	$4^{8} \cdot 2^{2} \cdot 5^{9}$
	↓	Write $2^{2} = 2 \cdot 2 = 4 = 4^{1}$ .
	$=$	$4^{8} \cdot 4^{1} \cdot 5^{9}$
	↓	Apply the power law $a^{x} \cdot a^{y} = a^{x + y}$ to $4^{8} \cdot 4^{1}$ .
	$=$	$4^{9} \cdot 5^{9}$
	↓	Apply the power law $a^{x} \cdot b^{x} = (a \cdot b)^{x}$ .
	$=$	$20^{9}$

${(7^{7})}^{7}$	$=$
	↓	Apply the power law ${(a^{x})}^{y} = a^{x \cdot y}$ .
	$=$	$7^{7 \cdot 7}$
	↓	Conclude the exponent
	$=$	$7^{49}$

$\frac{9^{2}}{9^{- 3}} : 3^{5}$	$=$
	↓	Apply the power law $\frac{a^{x}}{a^{y}} = a^{x - y}$ to $\frac{9^{2}}{9^{- 3}}$ .
	$=$	$9^{2 - (- 3)} : 3^{5}$
	↓	Summarize the exponent of $9$ .
	$=$	$9^{5} : 3^{5}$
	↓	Write $a : b = \frac{a}{b}$ .
	$=$	$\frac{9^{5}}{3^{5}}$
	↓	Use the power law $\frac{a^{x}}{b^{x}} = {(\frac{a}{b})}^{x}$ .
	$=$	${(\frac{9}{3})}^{5}$
	↓	Conclude the base.
	$=$	$3^{5}$

$\frac{\frac{26^{2}}{26^{8}}}{\frac{13^{- 3}}{13^{3}}}$	$=$
	↓	Use the power law $\frac{a^{x}}{a^{y}} = a^{x - y}$ with base $a = 26$ .
	$=$	$\frac{26^{- 6}}{\frac{13^{- 3}}{13^{3}}}$
	↓	Use the power law $\frac{a^{x}}{a^{y}} = a^{x - y}$ with base $a = 13$ .
	$=$	$\frac{26^{- 6}}{13^{- 6}}$
	↓	Use the power law $\frac{a^{x}}{b^{x}} = {(\frac{a}{b})}^{x}$ .
	$=$	${(\frac{26}{13})}^{- 6}$
	↓	Conclude the base.
	$=$	$2^{- 6}$