Exercises: Power Laws

Apply the power laws to simplify the following expressions:

$3^{2} \cdot 3^{1}$

For this task you need the following basic knowledge: Power Laws
$3^{2} \cdot 3^{1}$ $=$
↓
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.
$=$ $3^{3}$

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

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

${(7^{7})}^{7}$

For this task you need the following basic knowledge: Power Laws
${(7^{7})}^{7}$ $=$
↓
Apply the power law ${(a^{x})}^{y} = a^{x \cdot y}$ .
$=$ $7^{7 \cdot 7}$
↓
Conclude the exponent
$=$ $7^{49}$
$\frac{9^{2}}{9^{- 3}} : 3^{5}$

For this task you need the following basic knowledge: Power Laws
$\frac{9^{2}}{9^{- 3}} : 3^{5}$ $=$
↓
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.
$=$ $3^{5}$

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

2
Conclude as far as possible.
$a^{3} : a^{6}$

For this task you need the following basic knowledge: Powers
First Representation
$\frac{a \cdot a \cdot a}{a \cdot a \cdot a \cdot a \cdot a \cdot a}$ $=$
↓
Shorten by cancelling three factors of $a$ .
$=$ $\frac{1}{a \cdot a \cdot a}$
$=$ $\frac{1}{a^{3}} = a^{- 3}$
Second Representation
$a^{3} : a^{6}$ $=$
↓
Apply the power laws.
$=$ $a^{3 - 6}$
$=$ $a^{- 3}$

$2 x^{- 2} \cdot 3 x^{3}$

For this task you need the following basic knowledge: Powers
Power laws
$2 x^{- 2} \cdot 3 x^{3}$ $=$
↓
Use the commutative law to group numbers and variables together.
$=$ $2 \cdot 3 \cdot x^{- 2} \cdot x^{3}$
↓
Apply the power laws.
$=$ $2 \cdot 3 \cdot x^{- 2 + 3}$
$=$ $6 x$

$10^{- 12} : 10^{- 3}$

For this task you need the following basic knowledge: Powers
Power laws
$10^{- 12} : 10^{- 3}$ $=$
↓
Apply the power laws.
$=$ $10^{- 12 - (- 3)}$
$=$ $10^{- 9}$

$6 : 2^{3} - 9 \cdot 3^{- 2}$

For this task you need the following basic knowledge: Powers
$6 : 2^{3} - 9 \cdot 3^{- 2}$ $=$
↓
Write $9$ as $3^{2}$ .
$=$ $6 : 2^{3} - 3^{2} \cdot 3^{- 2}$
↓
Apply the power laws.
$=$ $6 : 8 - 3^{2 + (- 2)}$
$=$ $6 : 8 - 3^{0}$
$=$ $0.75 - 1$
$=$ $- 0.25$

$x^{- n} \cdot x$

For this task you need the following basic knowledge: Powers
$x^{- n} \cdot x$ $=$
↓
Write $x$ as $x^{1}$ .
$=$ $x^{- n} \cdot x^{1}$
↓
Apply the power laws.
$=$ $x^{- n + 1}$
Alternative solution
$x^{- n} \cdot x$ $=$
↓
Write $x^{- n}$ as $\frac{1}{x^{n}}$ .
$=$ $\frac{1}{x^{n}} \cdot x^{1}$
$=$ $\frac{x^{1}}{x^{n}}$
↓
Apply the power laws.
$=$ $x^{1 - n}$

$0.5 x^{2} + 1.5 x^{3}$

For this task you need the following basic knowledge: Powers
This term cannot be simplified further, since two different powers occur. However, what one could do is factorizing the term:
$0.5 x^{2} + 1.5 x^{3}$ $=$
↓
Factorize $0.5 x^{2}$ .
$=$ $0.5 x^{2} (1 + 3 x)$

${(\frac{x^{3} y^{- 4}}{y^{- 5} y^{2}})}^{- 2}$

For this task you need the following basic knowledge: Powers
First apply the power laws. You may get the minus out of the exponent by "flipping" the fractions like $x^{- 2} = \frac{1}{x^{2}}$ .
${(\frac{x^{3} y^{- 4}}{y^{- 5} y^{2}})}^{- 2}$ $=$ ${(\frac{y^{- 5} y^{2}}{x^{3} y^{- 4}})}^{2}$
↓
Apply the power laws.
$=$ ${(\frac{y^{- 5 + 2}}{x^{3} y^{- 4}})}^{2}$
$=$ ${(\frac{y^{- 3}}{x^{3} y^{- 4}})}^{2}$
↓
Shorten by $y^{- 3}$ .
$=$ ${(\frac{1}{x^{3} y^{- 1}})}^{2}$
↓
Apply the power laws once more.
$=$ ${(\frac{y}{x^{3}})}^{2}$
$=$ $\frac{y^{2}}{x^{6}}$

${(2 x^{3})}^{2}$

For this task you need the following basic knowledge: Powers
${(2 x^{3})}^{2}$ $=$
↓
Apply the power law ${(a \cdot b)}^{x} = a^{x} \cdot b^{x}$
$=$ $2^{2} \cdot x^{3 \cdot 2}$
$=$ $4 \cdot x^{6}$
3
Find all terms that are equivalent to each other:
Term 1: $x^{10}$
Term 2: $x^{- 6}$
Term 3: ${(x^{- 2})}^{4}$
Term 4: $x^{5} + x^{5}$
Term 5: ${(- x)}^{6}$
Term 6: $x^{- 8}$
Term 7: $x^{15} : x^{5}$
Term 8: $x^{- 22} \cdot x^{16}$
Term 9: $- x^{6}$

For this task you need the following basic knowledge: Power Laws
Term 1:
$x^{10}$
Term 2:
$x^{- 6}$
Term 3:
${(x^{- 2})}^{4} = x^{- 2 \cdot 4} = x^{- 8}$
Apply the power laws.
Term 4:
$x^{5} + x^{5} = 2 x^{5}$
Term 5:
${(- x)}^{6} = x^{6}$
The minus can be omitted here, because the power is even and $(- 1)^{2} = 1$ .
Term 6:
$x^{- 8}$
Term 7:
$x^{15} : x^{5} = x^{15 - 5} = x^{10}$
Apply the power laws.
Term 8:
$x^{- 22} \cdot x^{16} = x^{- 22 + 16} = x^{- 6}$
Apply the power laws.
Term 9:
$- x^{6} = - (x)^{6} = - x^{6}$
Equivalent terms:
Term 1 and term 7 are both equal to $x^{10}$ .
Term 2 and term 8 are both equal to $x^{6}$ .
Term 3 and term 6 are both equal to $x^{- 8}$ .

Simplify the following terms.

$10 \cdot 10^{- 2} : 10^{4} + 10^{0}$

$x^{- 1} \cdot x^{2} \cdot x^{0} \cdot x^{- 3} \cdot x^{4}$

For this task you need the following basic knowledge: Power Laws
$x^{- 1} \cdot x^{2} \cdot x^{0} \cdot x^{- 3} \cdot x^{4}$ $=$
↓
Apply the power laws.
$=$ $x^{- 1 + 2 + 0 - 3 + 4}$
$=$ $x^{2}$
$10^{- 1} + 10^{- 2}$

For this task you need the following basic knowledge: Power Laws
$10^{- 1} + 10^{- 2}$ $=$
↓
Convert into fractions.
$=$ $\frac{1}{10} + \frac{1}{100}$
↓
Get both fractions to a common denominator.
$=$ $\frac{10}{100} + \frac{1}{100}$
$=$ $\frac{11}{100}$
$x^{- 1} + x^{- 2}$

For this task you need the following basic knowledge: Power Laws
$x^{- 1} + x^{- 2}$ $=$
↓
Apply the power laws.
$=$ $\frac{1}{x} + \frac{1}{x^{2}}$
↓
Get both fractions to a common denominator.
$=$ $\frac{x}{x^{2}} + \frac{1}{x^{2}}$
$=$ $\frac{1 + x}{x^{2}}$
$x^{- 2} - \frac{x^{2}}{x^{4}}$

For this task you need the following basic knowledge: Power Laws
$x^{- 2} - \frac{x^{2}}{x^{4}}$ $=$
↓
Apply the power laws.
$=$ $x^{- 2} - x^{2 - 4}$
$=$ $x^{- 2} - x^{- 2}$
$=$ $0$
$(\frac{1}{x} + x^{- 2}) \cdot 2 x$

For this task you need the following basic knowledge: Power Laws
$(\frac{1}{x} + x^{- 2}) \cdot 2 x$ $=$
↓
Write as fractions.
$=$ $(\frac{1}{x} + \frac{1}{x^{2}}) \cdot 2 x$
↓
Multiply out.
$=$ $\frac{2 x}{x} + \frac{2 x}{x^{2}}$
↓
You can still shorten these fractions.
$=$ $\frac{2 \cdot x}{x} + \frac{2 \cdot x}{x \cdot x}$
$=$ $2 + \frac{2}{x}$

5
Simplify the following term using the power laws

$a^{4} \cdot d^{- 2} \cdot c^{9} \cdot a^{2} \cdot b^{6} \cdot d^{- 9} \cdot c^{5}$

For this task you need the following basic knowledge: Powers
$a^{4} \cdot d^{- 2} \cdot c^{9} \cdot a^{2} \cdot b^{6} \cdot d^{- 9} \cdot c^{5}$ $=$
↓
Conclude identical variables by applying the power laws.
$=$ $a^{4 + 2} \cdot b^{6} \cdot c^{9 + 5} \cdot d^{(- 2) + (- 9)}$
$=$ $a^{6} \cdot b^{6} \cdot c^{14} \cdot d^{- 11}$
↓
Conclude.
$=$ ${(a b)}^{6} \cdot c^{14} \cdot d^{- 11}$

Simplify the following expressions with integer exponents as far as possible.

$(z^{2 k - 5} : z^{3}) : z^{k}$

For this task you need the following basic knowledge: Powers
Apply the power laws.
$(z^{2 k - 5} : z^{3}) : z^{k}$ $=$ $(z^{2 k - 5 - 3}) : z^{k}$
$=$ $z^{2 k - 8} : z^{k}$
$=$ $z^{2 k - 8 - k}$
$=$ $z^{k - 8}$
$90 \cdot 3^{n - 2} - 3^{n}$

For this task you need the following basic knowledge: Powers
Write $90$ as $3^{2} \cdot 10$ .
$90 \cdot 3^{n - 2} - 3^{n}$ $=$ $10 \cdot 3^{2} \cdot 3^{n - 2} - 3^{n}$
↓
Apply the power laws.
$=$ $10 \cdot 3^{2 + n - 2} - 3^{n}$
$=$ $10 \cdot 3^{n} - 3^{n}$
↓
Factorize $3^{n}$ .
$=$ $3^{n} \cdot (10 - 1)$
$=$ $9 \cdot 3^{n}$
↓
Write $9$ as $3^{2}$ .
$=$ $3^{2} \cdot 3^{n}$
↓
Apply the power laws.
$=$ $3^{2 + n}$
${[{(\frac{x}{4})}^{3}]}^{5} : {(\frac{x}{2})}^{6}$ for $x \neq 0$

For this task you need the following basic knowledge: Powers
Apply the power laws.
${[{(\frac{x}{4})}^{3}]}^{5} : {(\frac{x}{2})}^{6}$ $=$ ${(\frac{x}{4})}^{3 \cdot 5} : (\frac{x^{6}}{2^{6}})$
$=$ ${(\frac{x}{4})}^{15} : (\frac{x^{6}}{2^{6}})$
$=$ $(\frac{x^{15}}{4^{15}}) : (\frac{x^{6}}{2^{6}})$
$=$ $(\frac{x^{15}}{4^{15}}) \cdot (\frac{2^{6}}{x^{6}})$
↓
Write $2^{6}$ as $4^{3}$ , so you can shorten.
$=$ $(\frac{x^{15}}{4^{15}}) \cdot (\frac{4^{3}}{x^{6}})$
$=$ $\frac{x^{9}}{4^{12}}$

$\frac{{(3 a - 1)}^{2 k - 1}}{{(1 - 3 a)}^{2 k + 1}}$ for $a \neq \frac{1}{3}$

${(\frac{6 a^{2} b^{- 2}}{c^{n + 1} d^{2 n}})}^{3} : {[\frac{2 {(cd)}^{n}}{{ab}^{- 1}} \cdot \frac{c^{n} d^{2 n}}{3 {ab}^{- 2}}]}^{- 2}$ for $a, b, c, d \neq 0$

\frac{x^{2 a + 5}}{{(- y^{3})}^{2 b + 5} \cdot {[{(- z)}^{4}]}^{3 b + 3}} : \frac{x^{2 a}}{{(yz)}^{6 b + 10} \cdot {[{(- z)}^{3}]}^{2 b - 1}}

Assuming that $x, y, z > 0$ , $b \in ℤ$

${(\frac{2 a^{- 1} b^{2}}{3 {ac}^{- 2}})}^{- 3}$ for $a, b, c \neq 0$

For this task you need the following basic knowledge: Powers
Apply the power laws.
${(\frac{2 a^{- 1} b^{2}}{3 {ac}^{- 2}})}^{- 3}$ $=$ $\frac{2^{- 3} a^{3} b^{- 6}}{3^{- 3} a^{- 3} c^{6}}$
$=$ $\frac{a^{3} 3^{3} a^{3}}{2^{3} c^{6} b^{6}}$
$=$ $\frac{a^{3} \cdot 27 a^{3}}{8 c^{6} b^{6}}$
$=$ $\frac{27 a^{6}}{8 b^{6} c^{6}}$

${(\frac{u}{v})}^{n} \cdot {(\frac{v}{u})}^{3 n + 4} : {(\frac{- v}{u})}^{2 n + 1}$ for $u, v \neq 0$

$\frac{x^{5} + 1}{x^{m + 2}} - \frac{2 x^{2} - 2}{x^{m}} + \frac{2 - x}{x^{m - 2}}$ for $x \neq 0$

${(\frac{z - 3}{z + 5})}^{2 p + 1} \cdot {(\frac{5 + z}{z - 3})}^{p + 1} : {(\frac{z - 3}{z + 5})}^{4 p}$ for $z \in̸ {- 5; 3}$

${(1 + \frac{2}{t})}^{2} \cdot {[\frac{1}{t} - {(\frac{t}{2} - 1)}^{- 1}]}^{- 2}$ for $t \in̸ {- 2; 0; 2}$

Re-formulate the expression to a single fraction that does not contain any negative exponents.

$\frac{4 a^{- 1} z^{2}}{{(x^{2} y)}^{3}} : \frac{{(2 a)}^{- 3}}{{({xy}^{2} z)}^{- 2}}$

7
Write as a decimal number.
$3 \cdot 10^{7}$

For this task you need the following basic knowledge: Powers
$3 \cdot 10^{7} = 3 \cdot 10 000 000 = 30 000 000$

$6.4 \cdot 10^{- 4}$

For this task you need the following basic knowledge: Powers
$6.4 \cdot 10^{- 4} = 6.4 \cdot 0.0001 = 0.00064$

$1.6 \cdot 10^{- 6}$

For this task you need the following basic knowledge: Powers
$1.6 \cdot 10^{- 6} = 1.6 \cdot 0.000001 = 0.0000016$

$7.4 \cdot 10^{9}$

For this task you need the following basic knowledge: Powers
$7.4 \cdot 10^{9} = 7.4 \cdot 1 000 000 000 = 7 400 000 000$
8
We are looking for powers with negative or positive exponents. Mark all correct answers with a cross.
$35 000 000 000$
$35 \cdot 10^{10}$
$3.5 \cdot 10^{- 8}$
$3.5 \cdot 10^{10}$
$3.5 \cdot 10^{9}$

For this task you need the following basic knowledge: Powers
$35 000 000 000 = 35 \cdot 1 000 000 000 = 35 \cdot 10^{9} = 3.5 \cdot 10^{10}$

$470 000 000$
$47 \cdot 10^{6}$
$47 \cdot 10^{7}$
$47 \cdot 10^{8}$

For this task you need the following basic knowledge: Powers
$470 000 000 = 47 \cdot 10 000 000 = 47 \cdot 10^{7}$

$0.0000001$
$10^{- 6}$
$10^{- 7}$
$10^{7}$
$10^{6}$

For this task you need the following basic knowledge: Powers
$0.0000001 = 10^{- 7}$

$0.0000054$
$5.4 \cdot 10^{- 5}$
$5.4 \cdot 10^{- 6}$
$5.4 \cdot 10^{- 7}$
$54 \cdot 10^{- 7}$

For this task you need the following basic knowledge: Powers
There are two solutions:
$0.0000054 = 5.4 \cdot 10^{- 6} = 54 \cdot 10^{- 7}$
9
Atoms are everywhere
A helium atom has a diameter of about $6 \cdot 10^{- 11}$ meters, and a hydrogen atom weighs about $1.7 \cdot 10^{- 27}$ kilograms.
The mass of Jupiter is about $1.899 \cdot 10^{27} k g$ , of which about $1.7 \cdot 10^{27}$ is hydrogen.
What idea can you get of the size of atoms and their mass?

For this task you need the following basic knowledge: Powers
Ein Wasserstoffatom hat ungefähr den Durchmesser von $6 \cdot 10^{- 11} m$ .
Die Zehnerpotenz kann man auch als Bruch schreiben. Das sieht dann so aus:
$10^{- 11} = \frac{1}{100 000 000 000}$ also sind $6 \cdot 10^{- 11} m$ sechs einhundert Milliardstel Meter.
Stell dir vor, du würdest einen Millimeter auf dem Lineal nochmal in $100$ Millionen Teile teilen. Davon nimmt man $6$ Teile. Dann ist man bei der Größe eines Atoms.
Die Masse des Atoms beträgt $1.7 \cdot 10^{- 27} k g$ .
Jetzt steht im Nenner des Bruchs eine $1$ mit $27$ Nullen. Diese Zahl nennt man auch Quadrilliarde. Ein Atom wiegt also ungefähr ein quadrilliardstel Kilogramm. Stell dir vor du nimmst einen gestrichenen Teelöffel mit Backpulver.
Dieser wiegt etwa $3$ g, also wiegt ein halber Teelöffel etwa $1.5$ g. Dieses kleine Häufchen Backpulver teilst du in eine Billionen Häufchen. Aber damit nicht getan. Eines dieser Häufchen teilst du nochmals in eine Billionen kleinere Häufchen. Die Masse eines dieser zwei Mal geteilten entspricht in etwa der Masse eines Wasserstoffatoms. Das ist eine unvorstellbar kleine Zahl!

Calculate the number of hydrogen atoms that Jupiter contains.

For this task you need the following basic knowledge: Powers
Die Masse des Wasserstoffanteils des Jupiters beträgt $m_{J} = 1.7 \cdot 10^{27} k g$ und die Masse eines Wasserstoffatoms: $m_{H} = 1.7 \cdot 10^{- 27} k g$ .
Teilt man nun die Masse des Wasserstoffanteils im Jupiter durch die Masse eines Wasserstoffatoms, so erhält man die Anzahl an Atomen.
$\frac{m_{J}}{m_{H}} = \frac{1.7 \cdot 10^{27} k g}{1.7 \cdot 10^{- 27} k g}$
Aus diesem Bruch kann man die Einheit kg und die $1,7$ direkt kürzen, da sie als Produkt im Zähler und im Nenner stehen.
$\frac{m_{J}}{m_{H}} = \frac{1.7 \cdot 10^{27} k g}{1.7 \cdot 10^{- 27} k g} = \frac{10^{27}}{10^{- 27}}$
Wende nun das Potenzgesetz zum Dividieren von Potenzen mit gleicher Basis an
$\frac{m_{J}}{m_{H}} = \frac{1.7 \cdot 10^{27} k g}{1.7 \cdot 10^{- 27} k g} = \frac{10^{27}}{10^{- 27}} = 10^{27 - (- 27)} = 10^{54}$
Der Jupiter enthält also die unglaublich hohe Anzahl von $10^{54}$ Wasserstoffatomen!
Diese Zahl nennt man übrigens Nonillion.

$3^{2} \cdot 3^{1}$	$=$
	↓	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.
	$=$	$2^{- 6}$

$\frac{a \cdot a \cdot a}{a \cdot a \cdot a \cdot a \cdot a \cdot a}$	$=$
	↓	Shorten by cancelling three factors of $a$ .
	$=$	$\frac{1}{a \cdot a \cdot a}$
	$=$	$\frac{1}{a^{3}} = a^{- 3}$

$a^{3} : a^{6}$	$=$
	↓	Apply the power laws.
	$=$	$a^{3 - 6}$
	$=$	$a^{- 3}$

$2 x^{- 2} \cdot 3 x^{3}$	$=$
	↓	Use the commutative law to group numbers and variables together.
	$=$	$2 \cdot 3 \cdot x^{- 2} \cdot x^{3}$
	↓	Apply the power laws.
	$=$	$2 \cdot 3 \cdot x^{- 2 + 3}$
	$=$	$6 x$

$10^{- 12} : 10^{- 3}$	$=$
	↓	Apply the power laws.
	$=$	$10^{- 12 - (- 3)}$
	$=$	$10^{- 9}$

$0.5 x^{2} + 1.5 x^{3}$	$=$
	↓	Factorize $0.5 x^{2}$ .
	$=$	$0.5 x^{2} (1 + 3 x)$

${(\frac{x^{3} y^{- 4}}{y^{- 5} y^{2}})}^{- 2}$	$=$	${(\frac{y^{- 5} y^{2}}{x^{3} y^{- 4}})}^{2}$
	↓	Apply the power laws.
	$=$	${(\frac{y^{- 5 + 2}}{x^{3} y^{- 4}})}^{2}$
	$=$	${(\frac{y^{- 3}}{x^{3} y^{- 4}})}^{2}$
	↓	Shorten by $y^{- 3}$ .
	$=$	${(\frac{1}{x^{3} y^{- 1}})}^{2}$
	↓	Apply the power laws once more.
	$=$	${(\frac{y}{x^{3}})}^{2}$
	$=$	$\frac{y^{2}}{x^{6}}$

$10^{- 1} + 10^{- 2}$	$=$
	↓	Convert into fractions.
	$=$	$\frac{1}{10} + \frac{1}{100}$
	↓	Get both fractions to a common denominator.
	$=$	$\frac{10}{100} + \frac{1}{100}$
	$=$	$\frac{11}{100}$

$x^{- 1} + x^{- 2}$	$=$
	↓	Apply the power laws.
	$=$	$\frac{1}{x} + \frac{1}{x^{2}}$
	↓	Get both fractions to a common denominator.
	$=$	$\frac{x}{x^{2}} + \frac{1}{x^{2}}$
	$=$	$\frac{1 + x}{x^{2}}$

$6 : 2^{3} - 9 \cdot 3^{- 2}$	$=$
	↓	Write $9$ as $3^{2}$ .
	$=$	$6 : 2^{3} - 3^{2} \cdot 3^{- 2}$
	↓	Apply the power laws.
	$=$	$6 : 8 - 3^{2 + (- 2)}$
	$=$	$6 : 8 - 3^{0}$
	$=$	$0.75 - 1$
	$=$	$- 0.25$

$x^{- n} \cdot x$	$=$
	↓	Write $x$ as $x^{1}$ .
	$=$	$x^{- n} \cdot x^{1}$
	↓	Apply the power laws.
	$=$	$x^{- n + 1}$

$x^{- n} \cdot x$	$=$
	↓	Write $x^{- n}$ as $\frac{1}{x^{n}}$ .
	$=$	$\frac{1}{x^{n}} \cdot x^{1}$
	$=$	$\frac{x^{1}}{x^{n}}$
	↓	Apply the power laws.
	$=$	$x^{1 - n}$

${(2 x^{3})}^{2}$	$=$
	↓	Apply the power law ${(a \cdot b)}^{x} = a^{x} \cdot b^{x}$
	$=$	$2^{2} \cdot x^{3 \cdot 2}$
	$=$	$4 \cdot x^{6}$

$10 \cdot 10^{- 2} : 10^{4} + 10^{0}$	$=$
	↓	Apply the power laws step by step on the left three terms.
	$=$	$10^{1 - 2} : 10^{4} + 10^{0}$
	$=$	$10^{- 1} : 10^{4} + 1$
	$=$	$10^{- 1 - 4} + 1$
	$=$	$10^{- 5} + 1$
	↓	For the sum, there is no power law. But you can conveniently express everything as a decimal number.
	$=$	$1.00001$

$x^{- 1} \cdot x^{2} \cdot x^{0} \cdot x^{- 3} \cdot x^{4}$	$=$
	↓	Apply the power laws.
	$=$	$x^{- 1 + 2 + 0 - 3 + 4}$
	$=$	$x^{2}$

$x^{- 2} - \frac{x^{2}}{x^{4}}$	$=$
	↓	Apply the power laws.
	$=$	$x^{- 2} - x^{2 - 4}$
	$=$	$x^{- 2} - x^{- 2}$
	$=$	$0$

$(\frac{1}{x} + x^{- 2}) \cdot 2 x$	$=$
	↓	Write as fractions.
	$=$	$(\frac{1}{x} + \frac{1}{x^{2}}) \cdot 2 x$
	↓	Multiply out.
	$=$	$\frac{2 x}{x} + \frac{2 x}{x^{2}}$
	↓	You can still shorten these fractions.
	$=$	$\frac{2 \cdot x}{x} + \frac{2 \cdot x}{x \cdot x}$
	$=$	$2 + \frac{2}{x}$

$a^{4} \cdot d^{- 2} \cdot c^{9} \cdot a^{2} \cdot b^{6} \cdot d^{- 9} \cdot c^{5}$	$=$
	↓	Conclude identical variables by applying the power laws.
	$=$	$a^{4 + 2} \cdot b^{6} \cdot c^{9 + 5} \cdot d^{(- 2) + (- 9)}$
	$=$	$a^{6} \cdot b^{6} \cdot c^{14} \cdot d^{- 11}$
	↓	Conclude.
	$=$	${(a b)}^{6} \cdot c^{14} \cdot d^{- 11}$

$(z^{2 k - 5} : z^{3}) : z^{k}$	$=$	$(z^{2 k - 5 - 3}) : z^{k}$
	$=$	$z^{2 k - 8} : z^{k}$
	$=$	$z^{2 k - 8 - k}$
	$=$	$z^{k - 8}$

Exercises: Power Laws

First Representation

Second Representation

Power laws

Power laws

Atoms are everywhere

$90 \cdot 3^{n - 2} - 3^{n}$	$=$	$10 \cdot 3^{2} \cdot 3^{n - 2} - 3^{n}$
	↓	Apply the power laws.
	$=$	$10 \cdot 3^{2 + n - 2} - 3^{n}$
	$=$	$10 \cdot 3^{n} - 3^{n}$
	↓	Factorize $3^{n}$ .
	$=$	$3^{n} \cdot (10 - 1)$
	$=$	$9 \cdot 3^{n}$
	↓	Write $9$ as $3^{2}$ .
	$=$	$3^{2} \cdot 3^{n}$
	↓	Apply the power laws.
	$=$	$3^{2 + n}$

${[{(\frac{x}{4})}^{3}]}^{5} : {(\frac{x}{2})}^{6}$	$=$	${(\frac{x}{4})}^{3 \cdot 5} : (\frac{x^{6}}{2^{6}})$
	$=$	${(\frac{x}{4})}^{15} : (\frac{x^{6}}{2^{6}})$
	$=$	$(\frac{x^{15}}{4^{15}}) : (\frac{x^{6}}{2^{6}})$
	$=$	$(\frac{x^{15}}{4^{15}}) \cdot (\frac{2^{6}}{x^{6}})$
	↓	Write $2^{6}$ as $4^{3}$ , so you can shorten.
	$=$	$(\frac{x^{15}}{4^{15}}) \cdot (\frac{4^{3}}{x^{6}})$
	$=$	$\frac{x^{9}}{4^{12}}$

$\frac{{(3 a - 1)}^{2 k - 1}}{{(1 - 3 a)}^{2 k + 1}}$	$=$	$\frac{{(- 1)}^{2 k - 1} {(- 3 a + 1)}^{2 k - 1}}{{(- 3 a + 1)}^{2 k + 1}}$
	↓	Divide and apply the power laws.
	$=$	${(- 1)}^{2 k - 1} {(- 3 a + 1)}^{2 k - 1 - (2 k + 1)}$
	↓	Multiply out.
	$=$	${(- 1)}^{2 k - 1} {(- 3 a + 1)}^{2 k - 1 - 2 k - 1}$
	$=$	${(- 1)}^{2 k - 1} {(- 3 a + 1)}^{- 2}$
	↓	A negative exponent corresponds to a fraction.
	$=$	${(- 1)}^{2 k - 1} \cdot \frac{1}{{(- 3 a + 1)}^{2}}$
	$=$	$\frac{{(- 1)}^{2 k - 1}}{{(- 3 a + 1)}^{2}}$
	$=$	$- \frac{1}{(- 3 a + 1)^{2}}$

${(\frac{6 a^{2} b^{- 2}}{c^{n + 1} d^{2 n}})}^{3} : {[\frac{2 {(cd)}^{n}}{{ab}^{- 1}} \cdot \frac{c^{n} d^{2 n}}{3 {ab}^{- 2}}]}^{- 2}$	$=$	$(\frac{6^{3} a^{6} b^{- 6}}{c^{3 n + 3} d^{6 n}}) : {(\frac{{ab}^{- 1} \cdot 3 {ab}^{- 2}}{2 {(cd)}^{n} \cdot c^{n} d^{2 n}})}^{2}$
	$=$	$(\frac{6^{3} a^{6}}{c^{3 n + 3} d^{6 n} \cdot b^{6}}) : (\frac{a^{2} b^{- 2} \cdot 3^{2} a^{2} b^{- 4}}{2^{2} {(cd)}^{2 n} \cdot c^{2 n} d^{4 n}})$
	↓	Convert division into multiplication by "flipping" the fraction.
	$=$	$(\frac{6^{3} a^{6}}{c^{3 n + 3} d^{6 n} \cdot b^{6}}) \cdot (\frac{2^{2} {(cd)}^{2 n} \cdot c^{2 n} d^{4 n}}{a^{2} b^{- 2} \cdot 3^{2} a^{2} b^{- 4}})$
	↓	Conclude.
	$=$	$\frac{216 a^{6} 4 c^{4 n} d^{6 n}}{9 a^{4} c^{3 n + 3} d^{6 n}}$
	$=$	$\frac{216 a^{2} 4 c^{n - 3}}{9}$
	$=$	$96 a^{2} c^{n - 3}$

$A$	$=$	$\frac{x^{2 a + 5}}{{(- y^{3})}^{2 b + 5} \cdot {[{(- z)}^{4}]}^{3 b + 3}} : \frac{x^{2 a}}{{(yz)}^{6 b + 10} \cdot {[{(- z)}^{3}]}^{2 b - 1}}$
	↓	Multiply out
	$=$	$\frac{x^{2 a + 5}}{{(- y^{3})}^{2 b + 5} \cdot {[{(- z)}^{4}]}^{3 b + 3}} \cdot \frac{{(yz)}^{6 b + 10} \cdot {[{(- z)}^{3}]}^{2 b - 1}}{x^{2 a}}$
	$=$	$\frac{x^{2 a + 5}}{{(- y^{3})}^{2 b + 5} \cdot {(- z)}^{4 (3 b + 3)}} \cdot \frac{{(yz)}^{6 b + 10} \cdot {(- z)}^{3 (2 b - 1)}}{x^{2 a}}$
	$=$	$\frac{x^{2 a + 5}}{{(- y)}^{6 b + 15} \cdot {(- z)}^{12 b + 12}} \cdot \frac{{(yz)}^{6 b + 10} \cdot {(- z)}^{6 b - 3}}{x^{2 a}}$
	↓	Factor out $(- 1)$ .
	$=$	$\frac{x^{2 a + 5}}{{(- 1)}^{6 b + 15} \cdot y^{6 b + 15} \cdot {(- 1)}^{12 b + 12} \cdot z^{12 b + 12}} \cdot \frac{{(yz)}^{6 b + 10} \cdot {(- 1)}^{6 b - 3} \cdot z^{6 b - 3}}{x^{2 a}}$
	↓	Decompose the $(y z)^{6 b + 10}$ .
	$=$	$\frac{x^{2 a + 5}}{{(- 1)}^{6 b + 15} \cdot y^{6 b + 15} \cdot {(- 1)}^{12 b + 12} \cdot z^{12 b + 12}} \cdot \frac{y^{6 b + 10} \cdot z^{6 b + 10} \cdot {(- 1)}^{6 b - 3} \cdot z^{6 b - 3}}{x^{2 a}}$
	$=$	$\frac{x^{2 a + 5 - 2 a}}{{(- 1)}^{6 b + 15} \cdot {(- 1)}^{12 b + 12}} \cdot y^{6 b + 10 - (6 b + 15)} \cdot z^{6 b + 10} \cdot {(- 1)}^{6 b - 3} \cdot z^{6 b - 3 - (12 b + 12)}$
	$=$	$\frac{x^{2 a + 5 - 2 a}}{{(- 1)}^{6 b + 15} \cdot {(- 1)}^{12 b + 12}} \cdot y^{6 b + 10 - 6 b - 15} \cdot z^{6 b + 10} \cdot {(- 1)}^{6 b - 3} \cdot z^{6 b - 3 - 12 b - 12}$
	↓	Resolve the bracket and simplify.
	$=$	$\frac{x^{5}}{{(- 1)}^{6 b + 15} \cdot {(- 1)}^{12 b + 12}} \cdot y^{- 5} \cdot z^{6 b + 10} \cdot {(- 1)}^{6 b - 3} \cdot z^{- 6 b - 15}$
	$=$	$\frac{x^{5} \cdot y^{- 5} \cdot z^{6 b + 10 - 6 b - 15} \cdot {(- 1)}^{6 b - 3}}{{(- 1)}^{6 b + 15} \cdot {(- 1)}^{12 b + 12}}$
	$=$	$\frac{x^{5} \cdot y^{- 5} \cdot z^{- 5} \cdot {(- 1)}^{6 b - 3}}{{(- 1)}^{6 b + 15} \cdot {(- 1)}^{12 b + 12}}$
	$=$	$\frac{x^{5} \cdot y^{- 5} \cdot z^{- 5} \cdot {(- 1)}^{6 b - 3}}{{(- 1)}^{6 b + 15 + 12 b + 12}}$
	$=$	$\frac{x^{5} \cdot y^{- 5} \cdot z^{- 5} \cdot {(- 1)}^{6 b - 3}}{{(- 1)}^{18 b + 27}}$
	↓	The factors of $(- 1)$ can be shortened.
	$=$	$\frac{x^{5} \cdot y^{- 5} \cdot z^{- 5} \cdot {(- 1)}^{6 b - 3 - (18 b + 27)}}{1}$
	$=$	$\frac{x^{5} \cdot y^{- 5} \cdot z^{- 5} \cdot {(- 1)}^{- 12 b - 30}}{1}$
	$=$	$x^{5} \cdot y^{- 5} \cdot z^{- 5} \cdot {(- 1)}^{- 12 b - 30}$
	$=$	$\frac{x^{5}}{y^{5} \cdot z^{5}} \cdot {(- 1)}^{- 12 b - 30}$

${(\frac{2 a^{- 1} b^{2}}{3 {ac}^{- 2}})}^{- 3}$	$=$	$\frac{2^{- 3} a^{3} b^{- 6}}{3^{- 3} a^{- 3} c^{6}}$
	$=$	$\frac{a^{3} 3^{3} a^{3}}{2^{3} c^{6} b^{6}}$
	$=$	$\frac{a^{3} \cdot 27 a^{3}}{8 c^{6} b^{6}}$
	$=$	$\frac{27 a^{6}}{8 b^{6} c^{6}}$

${(\frac{u}{v})}^{n} \cdot {(\frac{v}{u})}^{3 n + 4} : {(\frac{- v}{u})}^{2 n + 1}$	$=$	$\frac{u^{n}}{v^{n}} \cdot \frac{v^{3 n + 4}}{u^{3 n + 4}} : \frac{- v^{2 n + 1}}{u^{2 n + 1}}$
	↓	Write the division as a fraction.
	$=$	$\frac{\frac{u^{n}}{v^{n}} \cdot \frac{v^{3 n + 4}}{u^{3 n + 4}}}{\frac{- v^{2 n + 1}}{u^{2 n + 1}}}$
	↓	Resolve the double fraction.
	$=$	$\frac{u^{n}}{v^{n}} \cdot \frac{v^{3 n + 4}}{u^{3 n + 4}} \cdot \frac{u^{2 n + 1}}{- v^{2 n + 1}}$
	$=$	$\frac{u^{n} \cdot v^{3 n + 4} \cdot u^{2 n + 1}}{v^{n} \cdot u^{3 n + 4} \cdot (- v^{2 n + 1})}$
	$=$	$\frac{u^{n + 2 n + 1} \cdot v^{3 n + 4}}{u^{3 n + 4} \cdot (- 1) \cdot v^{n + 2 n + 1}}$
	$=$	$u^{3 n + 1 - (3 n + 4)} \cdot (- 1) \cdot v^{3 n + 4 - (3 n + 1)}$
	↓	Shorten by using the power laws.
	$=$	$u^{- 3} \cdot (- 1) \cdot v^{3}$
	$=$	$\frac{(- 1) \cdot v^{3}}{u^{3}}$
	$=$	$\frac{- v^{3}}{u^{3}}$
	$=$	$- {(\frac{v}{u})}^{3}$

$\frac{x^{5} + 1}{x^{m + 2}} - \frac{2 x^{2} - 2}{x^{m}} + \frac{2 - x}{x^{m - 2}}$	$=$	$\frac{x^{5} + 1}{x^{m + 2}} - \frac{2 x^{4} - 2 x^{2}}{x^{m + 2}} + \frac{2 - x}{x^{m} \cdot x^{- 2}}$
	↓	Apply the power laws to $x^{- 2}$ .
	$=$	$\frac{x^{5} + 1}{x^{m + 2}} - \frac{2 x^{4} - 2 x^{2}}{x^{m + 2}} + \frac{2 x^{2} - x^{3}}{x^{m}}$
	↓	Get to a common denominator.
	$=$	$\frac{x^{5} + 1}{x^{m + 2}} - \frac{2 x^{4} - 2 x^{2}}{x^{m + 2}} + \frac{2 x^{4} - x^{5}}{x^{m + 2}}$
	$=$	$\frac{x^{5} + 1 - 2 x^{4} + 2 x^{2} + 2 x^{4} - x^{5}}{x^{m + 2}}$
	$=$	$\frac{1 + 2 x^{2}}{x^{m + 2}}$

${(\frac{z - 3}{z + 5})}^{2 p + 1} \cdot {(\frac{5 + z}{z - 3})}^{p + 1} : {(\frac{z - 3}{z + 5})}^{4 p}$	$=$	${(\frac{z - 3}{z + 5})}^{2 p + 1 - 4 p} \cdot {(\frac{5 + z}{z - 3})}^{p + 1}$
	$=$	${(\frac{z - 3}{z + 5})}^{1 - 2 p} \cdot {(\frac{5 + z}{z - 3})}^{p + 1}$
	$=$	${(\frac{z - 3}{z + 5})}^{1 - 2 p} \cdot {(\frac{z - 3}{z + 5})}^{- (p + 1)}$
	↓	Resolve the bracket.
	$=$	${(\frac{z - 3}{z + 5})}^{1 - 2 p} \cdot {(\frac{z - 3}{z + 5})}^{- p - 1}$
	↓	Apply the power laws.
	$=$	${(\frac{z - 3}{z + 5})}^{1 - 2 p + (- p) - 1}$
	$=$	${(\frac{z - 3}{z + 5})}^{- 3 p}$
	$=$	${(\frac{z + 5}{z - 3})}^{3 p}$

$\frac{4 a^{- 1} z^{2}}{{(x^{2} y)}^{3}} : \frac{{(2 a)}^{- 3}}{{({xy}^{2} z)}^{- 2}}$	$=$	$\frac{4 \cdot \frac{1}{a} \cdot z^{2}}{x^{6} \cdot y^{3}} : \frac{\frac{1}{{(2 a)}^{3}}}{\frac{1}{{({xy}^{2} z)}^{2}}}$
	↓	Division can be done by "flipping the fraction".
	$=$	$\frac{4 \cdot \frac{1}{a} \cdot z^{2}}{x^{6} \cdot y^{3}} \cdot \frac{\frac{1}{{({xy}^{2} z)}^{- 2}}}{\frac{1}{{(2 a)}^{3}}}$
	↓	Resolve the double fraction.
	$=$	$\frac{4 \cdot \frac{1}{a} \cdot z^{2}}{x^{6} \cdot y^{3}} \cdot \frac{1}{{({xy}^{2} z)}^{2}} \cdot \frac{{(2 a)}^{3}}{1}$
	$=$	$\frac{\frac{4 z^{2}}{a}}{x^{6} \cdot y^{3}} \cdot \frac{8 a^{3}}{x^{2} \cdot y^{4} \cdot z^{2}}$
	$=$	$\frac{\frac{4 z^{2} \cdot 8 a^{3}}{a}}{x^{6 + 2} \cdot y^{3 + 4} \cdot z^{2}}$
	↓	Finally, shorten.
	$=$	$\frac{4 \cdot 8 a^{2}}{x^{8} y^{7}}$
	$=$	$\frac{32 a^{2}}{x^{8} y^{7}}$