Exercises: Power Laws

Apply the power laws to simplify the following expressions:

$3^2 \cdot 3^1$

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

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

$\left(7^7\right)^7$

$\dfrac{9^2}{9^{-3}}:3^{5}$

$\dfrac{\tfrac{26^2}{26^8}}{\tfrac{13^{-3}}{13^3}}$

Conclude as far as possible.

$a^3\;:\;a^6$

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

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

Für diese Aufgabe benötigst Du folgendes Grundwissen: Powers
Power laws
$\displaystyle 10^{-12}:10^{-3}$ $=$
↓
Apply the power laws.
$=$ $\displaystyle 10^{-12-\left(-3\right)}$
$=$ $\displaystyle 10^{-9}$
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$6\;:\;2^3-9\cdot3^{-2}$

$x^{-n}\;\cdot\;x$

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

Für diese Aufgabe benötigst Du folgendes Grundwissen: Powers
This term cannot be simplified further, since two different powers occur. However, what one could do is factorizing the term:
$\displaystyle 0.5x^2+1.5x^3$ $=$
↓
Factorize $0.5 x^{2}$ .
$=$ $\displaystyle 0.5x^2\left(1+3x\right)$
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$\left(\frac{x^3y^{-4}}{y^{-5}y^2}\right)^{-2}$

$\left(2x^3\right)^2$

3
Find all terms that are equivalent to each other:
Term 1: $x^{10}\;$
Term 2: $x^{- 6}$
Term 3: $\left(x^{-2}\right)^4$
Term 4: $x^{5} + x^{5}$
Term 5: $\left(-x\right)^6$
Term 6: $x^{- 8}$
Term 7: $x^{15}:x^5$
Term 8: $x^{-22}\cdot x^{16}$
Term 9: $- x^{6}$
Für diese Aufgabe benötigst Du folgendes Grundwissen: Power Laws
Term 1:
$x^{10}\;$
Term 2:
$x^{- 6}$
Term 3:
$\left(x^{-2}\right)^4=x^{-2\cdot4}=x^{-8}$
Apply the power laws.
Term 4:
$x^{5} + x^{5} = 2 x^{5}$
Term 5:
$\left(-x\right)^6=x^6$
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 $x^{6} x⁶$ .
Term 3 and term 6 are both equal to $x^{- 8}$ .

Simplify the following terms.

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

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

$10^{-1}+10^{-2}$

$x^{-1}+x^{-2}$

$x^{-2}-\frac{x^2}{x^4}$

$\left(\frac1x+x^{-2}\right)\cdot2x$

Simplify the following term using the power laws

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

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

$\left(z^{2k-5}:z^3\right)\;:\;z^k$

$90\cdot3^{n-2}-3^n$

$\left[\left(\frac x4\right)^3\right]^5:\;\left(\frac x2\right)^6$ for $x\neq 0$

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

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

Für diese Aufgabe benötigst Du folgendes Grundwissen: Powers

Apply the power laws.

$\displaystyle \left(\frac{6\mathrm a^2\mathrm b^{-2}}{\mathrm c^{\mathrm n+1}\mathrm d^{2\mathrm n}}\right)^3:\;\left[\frac{2\left(\mathrm{cd}\right)^\mathrm n}{\mathrm{ab}^{-1}}\;\cdot\;\frac{\mathrm c^\mathrm n\mathrm d^{2\mathrm n}}{3\mathrm{ab}^{-2}}\right]^{-2}$	$=$	$\displaystyle \left(\frac{6^3a^6b^{-6}}{c^{3n+3}d^{6n}}\right):\left(\frac{\mathrm{ab}^{-1}\cdot3\mathrm{ab}^{-2}}{2\left(\mathrm{cd}\right)^n\cdot c^nd^{2n}}\right)^2$
	$=$	$\displaystyle \left(\frac{6^3\mathrm a^6}{\mathrm c^{3\mathrm n+3}\mathrm d^{6\mathrm n}\cdot\mathrm b^6}\right):\left(\frac{\mathrm a^2\mathrm b^{-2}\cdot3^2\mathrm a^2\mathrm b^{-4}}{2^2\left(\mathrm{cd}\right)^{2\mathrm n}\cdot\mathrm c^{2\mathrm n}\mathrm d^{4\mathrm n}}\right)$
	↓	Convert division into multiplication by "flipping" the fraction.
	$=$	$\displaystyle \left(\frac{6^3\mathrm a^6}{\mathrm c^{3\mathrm n+3}\mathrm d^{6\mathrm n}\cdot\mathrm b^6}\right)\cdot\left(\frac{2^2\left(\mathrm{cd}\right)^{2\mathrm n}\cdot\mathrm c^{2\mathrm n}\mathrm d^{4\mathrm n}}{\mathrm a^2\mathrm b^{-2}\cdot3^2\mathrm a^2\mathrm b^{-4}}\right)$
	↓	Conclude.
	$=$	$\displaystyle \frac{216\mathrm a^64\mathrm c^{4\mathrm n}\mathrm d^{6\mathrm n}}{9\mathrm a^4\mathrm c^{3\mathrm n+3}\mathrm d^{6\mathrm n}}\;$
	$=$	$\displaystyle \frac{216\mathrm a^24\mathrm c^{\mathrm n-3}}9$
	$=$	$\displaystyle 96a^2c^{n-3}$

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

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

Für diese Aufgabe benötigst Du folgendes Grundwissen: Powers

Since $x$ is greater than $0$ , you can multiply by the reciprocal. For the value of

$A=\frac{x^{2a+5}}{\left(-y^3\right)^{2b+5}\cdot\;\left[\left(-z\right)^4\right]^{3b+3}}\;:\;\frac{x^{2a}}{\left(\mathrm{yz}\right)^{6b+10}\cdot\;\left[\left(-z\right)^3\right]^{2b-1}}$ you then obtain

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

Since $- 12 b - 30$ is even, you get $(-1)^{-12b-30}=1$ , so the final result reads $\frac{x^5}{y^5z^5}$ , or equivalently, $\left(\dfrac{x}{yz}\right)^5$ .

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

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

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

$\displaystyle\left(\frac{z-3}{z+5}\right)^{2p+1}\cdot\;\left(\frac{5+z}{z-3}\right)^{p+1}:\;\left(\frac{z-3}{z+5}\right)^{4p}$ for $z \not\in \{-5;3\}$

$\left(1+\frac2t\right)^2\cdot\;\left[\frac1t-\left(\frac t2-1\right)^{-1}\right]^{-2}$ for $t \not\in \{-2;0;2\}$

Für diese Aufgabe benötigst Du folgendes Grundwissen: Powers

Get the fraction in the rightmost round bracket on a common denominator.

$\displaystyle \left(1+\frac2t\right)^2\cdot\;\left[\frac1t-\left(\frac t2-1\right)^{-1}\right]^{-2}$	$=$	$\displaystyle \left(1+\frac2t\right)^2\cdot\;\left[\frac1t-\left(\frac{t-2}2\right)^{-1}\right]^{-2}$
	↓	Apply the power laws.
	$=$	$\displaystyle \left(1+\frac2t\right)^2\cdot\;\left[\frac1t-\left(\frac2{t-2}\right)\right]^{-2}$
	↓	Get the square bracket on a common denominator.
	$=$	$\displaystyle \left(1+\frac2t\right)^2\cdot\;\left[\frac{1\left(t-2\right)}{t\left(t-2\right)}-\frac{t\cdot2}{t\left(t-2\right)}\right]^{-2}$
	$=$	$\displaystyle \left(1+\frac2t\right)^2\cdot\;\left[\frac{t-2-2t}{t\left(t-2\right)}\right]^{-2}$
	$=$	$\displaystyle \left(1+\frac2t\right)^2\cdot\;\left[\frac{-t-2}{t\left(t-2\right)}\right]^{-2}$
	↓	Apply the power laws.
	$=$	$\displaystyle \left(1+\frac2t\right)^2\cdot\;\left[\frac{t\left(t-2\right)}{-t-2}\right]^2$
	↓	Get the round bracket on a common denominator.
	$=$	$\displaystyle \left(\frac tt+\frac2t\right)^2\cdot\;\left[\frac{t\left(t-2\right)}{-t-2}\right]^2$
	$=$	$\displaystyle \left(\frac{2+t}t\right)^2\cdot\;\left[\frac{t\left(t-2\right)}{-t-2}\right]^2$
	↓	Apply the power laws to conclude both brackets.
	$=$	$\displaystyle \left(\frac{\left(2+t\right)\cdot t\left(t-2\right)}{t\cdot\left(-t-2\right)}\right)^2$
	↓	Shorten by $t$ .
	$=$	$\displaystyle \left(\frac{\left(2+t\right)\cdot\left(t-2\right)}{\left(-t-2\right)}\right)^2$
	↓	Factor out a $(- 1)$ .
	$=$	$\displaystyle \left(\frac{\left(2+t\right)\cdot\left(t-2\right)}{-1\left(t+2\right)}\right)^2$
	↓	Shorten.
	$=$	$\displaystyle \left(\frac{t-2}{-1}\right)^2$
	$=$	$\displaystyle (-(t-2))^2$
	$=$	$\displaystyle \left(t-2\right)^2$

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

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

7
Write as a decimal number.
1. $3\cdot10^7$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Powers
  $3\cdot10^7\ =\ 3\cdot10\ 000\ 000\ =\ 30\ 000\ 000$
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2. $6.4\cdot10^{-4}$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Powers
  $6.4\cdot10^{-4}\ =\ 6.4\ \cdot0.0001\ =\ 0.00064$
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3. $1.6\cdot10^{-6}$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Powers
  $1.6\cdot10^{-6}\ =\ 1.6\ \cdot0.000001\ =\ 0.0000016$
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4. $7.4\cdot10^9$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Powers
  $7.4\cdot10^9\ =\ 7.4\cdot1\ 000\ 000\ 000\ =\ 7\ 400\ 000\ 000$
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8
We are looking for powers with negative or positive exponents. Mark all correct answers with a cross.
1. $35\ 000\ 000\ 000$
  $35\cdot10^{10}$
  $3.5\cdot10^{-8}$
  $3.5\cdot10^{10}$
  $3.5\cdot10^9$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Powers
  $35\ 000\ 000\ 000\ =\ 35\ \cdot1\ 000\ 000\ 000\ =\ 35\ \cdot10^9\ =\ 3.5\ \cdot10^{10}$
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2. $470\ 000\ 000$
  $47\cdot10^6$
  $47\cdot10^7$
  $47\cdot10^8$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Powers
  $470\ 000\ 000\ =\ 47\ \cdot10\ 000\ 000\ =\ 47\ \cdot10^7$
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  Write it here 👉
3. $0.0000001$
  $10^{-6}$
  $10^{-7}$
  $10^{7}$
  $10^{6}$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Powers
  $0.0000001=10^{-7}$
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4. $0.0000054$
  $5.4\cdot10^{-5}$
  $5.4\cdot10^{-6}$
  $5.4\cdot10^{-7}$
  $54\cdot10^{-7}$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Powers
  There are two solutions:
  $0.0000054 = 5.4 \cdot 10^{-6} = 54 \cdot 10^{-7}$
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9
Atoms are everywhere
A helium atom has a diameter of about $6\cdot10^{-11}$ meters, and a hydrogen atom weighs about $1.7 \cdot 10^{-27}$ kilograms.
The mass of Jupiter is about $1.899 \cdot 10^{27} kg$ , of which about $1.7 \cdot 10^{27}$ is hydrogen.
1. What idea can you get of the size of atoms and their mass?
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Powers
  Ein Wasserstoffatom hat ungefähr den Durchmesser von $6\ \cdot10^{-11}m$ .
  Die Zehnerpotenz kann man auch als Bruch schreiben. Das sieht dann so aus:
  $10^{-11\ }=\frac{1}{100\ 000\ 000\ 000}$ also sind $6\ \cdot10^{-11}m$ 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 $1.7\ \cdot10^{-27}kg$ .
  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!
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2. Calculate the number of hydrogen atoms that Jupiter contains.
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Powers
  Die Masse des Wasserstoffanteils des Jupiters beträgt $m_J=1.7\ \cdot10^{27\ }kg$ und die Masse eines Wasserstoffatoms: $m_H=1.7\ \cdot10^{-27}kg$ .
  Teilt man nun die Masse des Wasserstoffanteils im Jupiter durch die Masse eines Wasserstoffatoms, so erhält man die Anzahl an Atomen.
  $\displaystyle \frac{m_J}{m_H}=\frac{1.7\cdot10^{27}\ kg}{1.7\cdot10^{-27}\ kg}$
  Aus diesem Bruch kann man die Einheit kg und die $1,7$ direkt kürzen, da sie als Produkt im Zähler und im Nenner stehen.
  $\displaystyle \frac{m_J}{m_H}=\frac{1.7\cdot10^{27}kg}{1.7\cdot10^{-27}kg}=\frac{10^{27}}{10^{-27}}$
  Wende nun das Potenzgesetz zum Dividieren von Potenzen mit gleicher Basis an
  $\displaystyle \frac{m_J}{m_H}=\frac{1.7\cdot10^{27}\ kg}{1.7\cdot10^{-27}\ kg}=\frac{10^{27}}{10^{-27}}=10^{27-\left(-27\right)}=10^{54}$
  Der Jupiter enthält also die unglaublich hohe Anzahl von $10^{54}$ Wasserstoffatomen!
  Diese Zahl nennt man übrigens Nonillion.
  Do you have a question?
  Write it here 👉

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$\displaystyle \left(z^{2k-5}:z^3\right)\;:\;z^{k\;}$	$=$	$\displaystyle \left(z^{2k-5-3}\right):z^k$
	$=$	$\displaystyle z^{2k-8}:z^k$
	$=$	$\displaystyle z^{2k-8-k}$
	$=$	$\displaystyle z^{k-8}$

$\displaystyle \left[\left(\frac x4\right)^3\right]^5:\;\left(\frac x2\right)^6$	$=$	$\displaystyle \left(\frac x4\right)^{3\cdot5}:\left(\frac{x^6}{2^6}\right)$
	$=$	$\displaystyle \left(\frac x4\right)^{15}:\left(\frac{x^6}{2^6}\right)$
	$=$	$\displaystyle \left(\frac{x^{15}}{4^{15}}\right):\left(\frac{x^6}{2^6}\right)$
	$=$	$\displaystyle \left(\frac{x^{15}}{4^{15}}\right)\cdot\left(\frac{2^6}{x^6}\right)$
	↓	Write $2^{6}$ as $4^{3}$ , so you can shorten.
	$=$	$\displaystyle \left(\frac{x^{15}}{4^{15}}\right)\cdot\left(\frac{4^3}{x^6}\right)$
	$=$	$\displaystyle \frac{x^9}{4^{12}}$

$\displaystyle \frac{\left(3a-1\right)^{2k-1}}{\left(1-3a\right)^{2k+1}}$	$=$	$\displaystyle \frac{\left(-1\right)^{2k-1}\left(-3a+1\right)^{2k-1}}{\left(-3a+1\right)^{2k+1}}$
	↓	Divide and apply the power laws.
	$=$	$\displaystyle \left(-1\right)^{2k-1}\left(-3a+1\right)^{2k-1-\left(2k+1\right)}$
	↓	Multiply out.
	$=$	$\displaystyle \left(-1\right)^{2k-1}\left(-3a+1\right)^{2k-1-2k-1}$
	$=$	$\displaystyle \left(-1\right)^{2k-1}\left(-3a+1\right)^{-2}$
	↓	A negative exponent corresponds to a fraction.
	$=$	$\displaystyle \left(-1\right)^{2k-1}\cdot\frac1{\left(-3a+1\right)^2}$
	$=$	$\displaystyle \frac{\left(-1\right)^{2k-1}}{\left(-3a+1\right)^2}$
	$=$	$\displaystyle -\frac1{(-3a+1)^2}$

$\displaystyle \left(\frac{2a^{-1}b^2}{3\mathrm{ac}^{-2}}\right)^{-3}$	$=$	$\displaystyle \frac{2^{-3}a^3b^{-6}}{3^{-3}a^{-3}c^6}$
	$=$	$\displaystyle \frac{a^33^3a^3}{2^3c^6b^6}$
	$=$	$\displaystyle \frac{a^3\cdot27a^3}{8c^6b^6}$
	$=$	$\displaystyle \frac{27a^6}{8b^6c^6}$

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

$\displaystyle 3^2\cdot3^1$	$=$
	↓	Use the power law $a^x \cdot a^y = a^{x+y}$ with $x = 2, y = 1, a = 3$ .
	$=$	$\displaystyle 3^{2+1}$
	↓	Conclude the exponent.
	$=$	$\displaystyle 3^3$

$\displaystyle 4^2\cdot4^9\cdot4^{-12}$	$=$
	↓	First apply the power laws to $4^2 \cdot 4^9$ .
	$=$	$\displaystyle 4^{2+9}\cdot4^{-12}$
	$=$	$\displaystyle 4^{11}\cdot4^{-12}$
	↓	Now apply the power law to $4^{11} \cdot 4^{-12}$ .
	$=$	$\displaystyle 4^{-1}$
	↓	The term can be simplified even further by using the rule of the negative exponent, that is $a^{-1}=\frac{1}{a}.$
	$=$	$\displaystyle \frac{1}{4}$

$\displaystyle 4^8\cdot2^{-3}\cdot2^5\cdot5^9$	$=$
	↓	Apply the power law $a^{x} \cdot a^y = a^{x +y}$ to $2^{-3} \cdot 2^5$ .
	$=$	$\displaystyle 4^8\cdot2^2\cdot5^9$
	↓	Write $2^2=2\cdot2=4=4^1$ .
	$=$	$\displaystyle 4^8\cdot4^1\cdot5^9$
	↓	Apply the power law $a^{x} \cdot a^y = a^{x +y}$ to $4^8 \cdot 4^1$ .
	$=$	$\displaystyle 4^9\cdot5^9$
	↓	Apply the power law $a^x \cdot b^x = (a\cdot b)^x$ .
	$=$	$\displaystyle 20^9$

$\displaystyle \left(7^7\right)^7$	$=$
	↓	Apply the power law $\left(a^x \right)^y=a^{x\cdot y}$ .
	$=$	$\displaystyle 7^{7\cdot7}$
	↓	Conclude the exponent
	$=$	$\displaystyle 7^{49}$

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

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

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

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

$\displaystyle 2x^{-2}\cdot3x^3$	$=$
	↓	Use the commutative law to group numbers and variables together.
	$=$	$\displaystyle 2\cdot3\cdot x^{-2}\cdot x^3$
	↓	Apply the power laws.
	$=$	$\displaystyle 2\cdot3\cdot x^{-2+3}$
	$=$	$\displaystyle 6x$

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

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

$\displaystyle 6:2^3-9\cdot3^{-2}$	$=$
	↓	Write $9$ as $3^{2}$ .
	$=$	$\displaystyle 6:2^3-3^2\cdot3^{-2}$
	↓	Apply the power laws.
	$=$	$\displaystyle 6:8-3^{2+\left(-2\right)}$
	$=$	$\displaystyle 6:8-3^0$
	$=$	$\displaystyle 0.75-1$
	$=$	$\displaystyle -0.25$

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

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

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

$\displaystyle \left(2x^3\right)^2$	$=$
	↓	Apply the power law $\left(a\cdot b\right)^x=a^x\cdot b^x$
	$=$	$\displaystyle 2^2\cdot x^{3\cdot2}$
	$=$	$\displaystyle 4\cdot x^6$

$\displaystyle 10\cdot10^{-2}:10^4+10^0$	$=$
	↓	Apply the power laws step by step on the left three terms.
	$=$	$\displaystyle 10^{1-2}:10^4+10^0$
	$=$	$\displaystyle 10^{-1}:10^4+1$
	$=$	$\displaystyle 10^{-1-4}+1$
	$=$	$\displaystyle 10^{-5}+1$
	↓	For the sum, there is no power law. But you can conveniently express everything as a decimal number.
	$=$	$\displaystyle 1.00001$

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

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

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

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

Exercises: Power Laws

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First Representation

Second Representation

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Power laws

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Power laws

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Atoms are everywhere

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$\displaystyle \left(\frac{1}{x}+x^{-2}\right)\cdot2x$	$=$
	↓	Write as fractions.
	$=$	$\displaystyle \left(\frac{1}{x}+\frac{1}{x^2}\right)\cdot2x$
	↓	Multiply out.
	$=$	$\displaystyle \frac{2x}{x}+\frac{2x}{x^2}$
	↓	You can still shorten these fractions.
	$=$	$\displaystyle \frac{2\cdot x}{x}+\frac{2\cdot x}{x\cdot x}$
	$=$	$\displaystyle 2+\frac{2}{x}$

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

$\displaystyle 90\cdot3^{n-2}-3^n$	$=$	$\displaystyle 10\cdot3^2\cdot3^{n-2}-3^n$
	↓	Apply the power laws.
	$=$	$\displaystyle 10\cdot3^{2+n-2}-3^n$
	$=$	$\displaystyle 10\cdot3^n-3^n$
	↓	Factorize $3^{n}$ .
	$=$	$\displaystyle 3^n\cdot(10-1)$
	$=$	$\displaystyle 9\cdot3^n$
	↓	Write $9$ as $3^{2}$ .
	$=$	$\displaystyle 3^2\cdot3^n$
	↓	Apply the power laws.
	$=$	$\displaystyle 3^{2+n}$

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

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

$\displaystyle \frac{4a^{-1}z^2}{\left(x^2y\right)^3}\;:\;\frac{\left(2a\right)^{-3}}{\left(\mathrm{xy}^2z\right)^{-2}}$	$=$	$\displaystyle \frac{4\cdot\frac1a\cdot z^2}{x^6\cdot y^3}:\frac{\frac1{\left(2a\right)^3}}{\frac1{\left(\mathrm{xy}^2z\right)^2}}$
	↓	Division can be done by "flipping the fraction".
	$=$	$\displaystyle \frac{4\cdot\frac1a\cdot z^2}{x^6\cdot y^3}\cdot\frac{\frac1{\left(\mathrm{xy}^2z\right)^{-2}}}{\frac1{\left(2a\right)^3}}$
	↓	Resolve the double fraction.
	$=$	$\displaystyle \frac{4\cdot\frac1a\cdot z^2}{x^6\cdot y^3}\cdot\frac1{\left(\mathrm{xy}^2z\right)^2}\cdot\frac{\left(2a\right)^3}1$
	$=$	$\displaystyle \frac{\frac{4z^2}a}{x^6\cdot y^3}\cdot\frac{8a^3}{x^2\cdot y^4\cdot z^2}$
	$=$	$\displaystyle \frac{\frac{4z^2\cdot8a^3}a}{x^{6+2}\cdot y^{3+4}\cdot z^2}$
	↓	Finally, shorten.
	$=$	$\displaystyle \frac{4\cdot8a^2}{x^8y^7}$
	$=$	$\displaystyle \frac{32a^2}{x^8y^7}$