1.3 Powers - Revision history

Jdinglis at 13:57, 26 October 2008

Jdinglis — Sun, 26 Oct 2008 13:57:47 GMT

←Older revision		Revision as of 13:57, 26 October 2008
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-	'''Keep in mind that:'''	+	'''Keep in mind that...'''

	The number raised to the power 0 is always 1 as long as the number (the base) is not 0.		The number raised to the power 0 is always 1 as long as the number (the base) is not 0.

Eakc07: Minor Grammatical changes

Eakc07 — Wed, 22 Oct 2008 00:00:27 GMT

Minor Grammatical changes

←Older revision		Revision as of 00:00, 22 October 2008
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	<ol type="a">		<ol type="a">
-	<li><math>\quad 3^{5/6} > 3^{3/4}\quad</math> as the base <math>3</math> is greater than <math>1</math> and the first exponent <math>5/6</math> is greater than the second exponent <math>3/4</math>.</li>	+	<li><math>\quad 3^{5/6} > 3^{3/4}\quad</math> because the base <math>3</math> is greater than <math>1</math> and the first exponent <math>5/6</math> is greater than the second exponent <math>3/4</math>.</li>
	<li><math>\quad 3^{-3/4} > 3^{-5/6}\quad</math> as the base is greater than <math>1</math> and the exponents satisfy <math> -3/4 > - 5/6</math>.</li>		<li><math>\quad 3^{-3/4} > 3^{-5/6}\quad</math> as the base is greater than <math>1</math> and the exponents satisfy <math> -3/4 > - 5/6</math>.</li>
-	<li><math> \quad 0{,}3^5 < 0{,}3^4 \quad</math>as the base <math> 0{,}3</math> is between <math>0</math> and <math>1</math> and <math>5 > 4</math>.	+	<li><math> \quad 0\text{.}3^5 < 0\text{.}3^4 \quad</math>as the base <math> 0\text{.}3</math> is between <math>0</math> and <math>1</math> and <math>5 > 4</math>.
	</ol>		</ol>
	</div>		</div>

-	If a power has a positive exponent, it increases as the base increases. The opposite applies if the exponent is negative, that is the power decreases as the base increases.	+	If a power has a positive exponent it increases as the base increases. The opposite applies if the exponent is negative, that is to say the power decreases as the base increases.

	<div class="exempel">		<div class="exempel">
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	</div>		</div>

-	Sometimes powers need to be rewritten in order to determine the relative sizes. For example to compare <math>125^2</math> with <math>36^3</math>~~one~~ can rewrite them as	+	Sometimes powers need to be rewritten in order to determine the relative sizes. For example to compare <math>125^2</math> with <math>36^3</math> we can rewrite them as
	{{Displayed math\|\|<math>		{{Displayed math\|\|<math>
	125^2 = (5^3)^2 = 5^6\quad \text{and}\quad 36^3 = (6^2)^3 = 6^6		125^2 = (5^3)^2 = 5^6\quad \text{and}\quad 36^3 = (6^2)^3 = 6^6
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	''' Example 11'''		''' Example 11'''

-	Determine which of the following pairs of numbers is the greater	+	Determine which of the following pairs of numbers is the greater:

	<ol type="a">		<ol type="a">
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	The base 25 can be rewritten in terms of the second base <math>5</math> by putting <math>25= 5\cdot 5= 5^2</math>. Therefore		The base 25 can be rewritten in terms of the second base <math>5</math> by putting <math>25= 5\cdot 5= 5^2</math>. Therefore

-	{{Displayed math\|\|<math>25^{1/3} = (5^2)^{1/3} = 5^{2 \cdot \frac{1}{3}}= 5^{2/3}</math>}}	+	{{Displayed math\|\|<math>25^{1/3} = (5^2)^{1/3} = 5^{2 \cdot \frac{1}{3}}= 5^{2/3}\text{,}</math>}}

-	~~and then~~ we see that	+	hence we see that

	{{Displayed math\|\|<math>5^{3/4} > 25^{1/3} </math>}}		{{Displayed math\|\|<math>5^{3/4} > 25^{1/3} </math>}}
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	{{Displayed math\|\|<math>\eqalign{8 &= 2\cdot 4 = 2 \cdot 2 \cdot 2 = 2^3\mbox{,}\\ 128 &= 2\cdot 64 = 2\cdot 2\cdot 32 = 2\cdot 2\cdot 2\cdot 16 = 2\cdot 2\cdot 2\cdot 2\cdot 8\\ &= 2\cdot 2\cdot 2\cdot 2\cdot 2^3 = 2^7\mbox{.}}</math>}}		{{Displayed math\|\|<math>\eqalign{8 &= 2\cdot 4 = 2 \cdot 2 \cdot 2 = 2^3\mbox{,}\\ 128 &= 2\cdot 64 = 2\cdot 2\cdot 32 = 2\cdot 2\cdot 2\cdot 16 = 2\cdot 2\cdot 2\cdot 2\cdot 8\\ &= 2\cdot 2\cdot 2\cdot 2\cdot 2^3 = 2^7\mbox{.}}</math>}}

-	This ~~means that~~	+	This gives

	{{Displayed math\|\|<math>\begin{align*}		{{Displayed math\|\|<math>\begin{align*}
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	<br>		<br>
	<br>		<br>
-	Since <math>8=2^3</math> and <math>27=3^3</math> a first step ~~can be~~ to simplify and write the numbers as powers of <math>2</math> and <math>3</math> respectively,	+	Since <math>8=2^3</math> and <math>27=3^3</math>, the first step is to simplify and write the numbers as powers of <math>2</math> and <math>3</math> respectively,

	{{Displayed math\|\|<math>\begin{align*}		{{Displayed math\|\|<math>\begin{align*}
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	<br>		<br>
	<br>		<br>
-	We rewrite the exponents ~~so they have~~ a common denominator	+	We rewrite the exponents due to them having a common denominator

	{{Displayed math\|\|<math>\frac{1}{3} = \frac{2}{6} \quad</math> and <math>\quad \frac{1}{2} = \frac{3}{6}</math>.}}		{{Displayed math\|\|<math>\frac{1}{3} = \frac{2}{6} \quad</math> and <math>\quad \frac{1}{2} = \frac{3}{6}</math>.}}

-	~~Then we have that~~	+	This gives

	{{Displayed math\|\|<math>\begin{align*}		{{Displayed math\|\|<math>\begin{align*}
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	'''Keep in mind that:'''		'''Keep in mind that:'''

-	The number raised to the power 0, is always 1~~, if~~ the number (the base) is not 0.	+	The number raised to the power 0 is always 1 as long as the number (the base) is not 0.

Eakc07: Minor grammatical changes

Eakc07 — Tue, 21 Oct 2008 23:37:07 GMT

Minor grammatical changes

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Tekbot: Robot: Automated text replacement (-{{Fristående formel +{{Displayed math)

Tekbot — Wed, 10 Sep 2008 13:54:09 GMT

Robot: Automated text replacement (-{{Fristående formel +{{Displayed math)

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Tekbot: Robot: Automated text replacement (-{{Vald flik +{{Selected tab)

Tekbot — Wed, 10 Sep 2008 13:49:15 GMT

Robot: Automated text replacement (-{{Vald flik +{{Selected tab)

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-	{{~~Vald flik~~\|[[1.3 Powers\|Theory]]}}	+	{{Selected tab\|[[1.3 Powers\|Theory]]}}
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Tekbot: Robot: Automated text replacement (-{{Ej vald flik +{{Not selected tab)

Tekbot — Wed, 10 Sep 2008 13:43:52 GMT

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	{{Vald flik\|[[1.3 Powers\|Theory]]}}		{{Vald flik\|[[1.3 Powers\|Theory]]}}
-	{{~~Ej vald flik~~\|[[1.3 Exercises\|Exercises]]}}	+	{{Not selected tab\|[[1.3 Exercises\|Exercises]]}}
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Tekbot: Robot: Automated text replacement (-1.3 Potenser +1.3 Powers)

Tekbot — Wed, 10 Sep 2008 13:12:26 GMT

Robot: Automated text replacement (-1.3 Potenser +1.3 Powers)

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-	{{Vald flik\|[[1.3 ~~Potenser~~\|Theory]]}}	+	{{Vald flik\|[[1.3 Powers\|Theory]]}}
	{{Ej vald flik\|[[1.3 Exercises\|Exercises]]}}		{{Ej vald flik\|[[1.3 Exercises\|Exercises]]}}
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Tekbot: Robot: Automated text replacement (-Övningar +Exercises)

Tekbot — Wed, 10 Sep 2008 10:37:41 GMT

Robot: Automated text replacement (-Övningar +Exercises)

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	{{Vald flik\|[[1.3 Potenser\|Theory]]}}		{{Vald flik\|[[1.3 Potenser\|Theory]]}}
-	{{Ej vald flik\|[[1.3 ~~Övningar~~\|Exercises]]}}	+	{{Ej vald flik\|[[1.3 Exercises\|Exercises]]}}
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	</div>		</div>

-	[[1.3 ~~Övningar~~\|Exercises]]	+	[[1.3 Exercises\|Exercises]]

Tekbot: 1.3 Potenser moved to 1.3 Powers: Robot: moved page

Tekbot — Tue, 09 Sep 2008 12:18:53 GMT

1.3 Potenser moved to 1.3 Powers: Robot: moved page

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Tek: small fixes

Tek — Thu, 14 Aug 2008 12:07:22 GMT

small fixes

←Older revision		Revision as of 12:07, 14 August 2008
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	There are a few more rules coming from the definition of power which are useful when doing calculations.You can see for example that		There are a few more rules coming from the definition of power which are useful when doing calculations.You can see for example that

-	{{Fristående formel\|\|<math>2^3 \cdot 2^5 = \underbrace{\,2\cdot 2\cdot 2\vphantom{{}_{\scriptscriptstyle 1}}\,}_{ 3\ {\rm st }} \cdot \underbrace{\,2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\vphantom{{}_{\scriptscriptstyle 1}}\,}_{ 5\ {\rm st }} = \underbrace{\,2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\vphantom{{}_{\scriptscriptstyle 1}}\,}_{ (3 + 5)\ {\rm st}} = 2^{3+5} = 2^8</math>}}	+	{{Fristående formel\|\|<math>2^3 \cdot 2^5 = \underbrace{\,2\cdot 2\cdot 2\vphantom{{}_{\scriptscriptstyle 1}}\,}_{ 3\ {\rm factors }} \cdot \underbrace{\,2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\vphantom{{}_{\scriptscriptstyle 1}}\,}_{ 5\ {\rm factors }} = \underbrace{\,2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\vphantom{{}_{\scriptscriptstyle 1}}\,}_{ (3 + 5)\ {\rm factors}} = 2^{3+5} = 2^8</math>}}

	which generally can be expressed as		which generally can be expressed as
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	For the case when the base itself is a power one has another useful rule. We see that		For the case when the base itself is a power one has another useful rule. We see that

-	{{Fristående formel\|\|<math> (5^2)^3 = 5^2 \cdot 5^2 \cdot 5^2 = \underbrace{\,5\cdot 5\vphantom{{}_{\scriptscriptstyle 1}}\,}_{ 2\ {\rm st}} \cdot \underbrace{\,5\cdot 5\vphantom{{}_{\scriptscriptstyle 1}}\,}_{ 2\ {\rm st}} \cdot \underbrace{\,5\cdot 5\vphantom{{}_{\scriptscriptstyle 1}}\,}_{2\ {\rm st}} = \underbrace{\,5\cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5\vphantom{{}_{\scriptscriptstyle 1}}\,}_{3\ {\rm ~~gånger~~}\ 2\ {\rm st}} = 5^{2 \cdot 3} = 5^6\mbox{.}</math>}}	+	{{Fristående formel\|\|<math> (5^2)^3 = 5^2 \cdot 5^2 \cdot 5^2 = \underbrace{\,5\cdot 5\vphantom{{}_{\scriptscriptstyle 1}}\,}_{ 2\ {\rm factors}} \cdot \underbrace{\,5\cdot 5\vphantom{{}_{\scriptscriptstyle 1}}\,}_{ 2\ {\rm factors}} \cdot \underbrace{\,5\cdot 5\vphantom{{}_{\scriptscriptstyle 1}}\,}_{2\ {\rm factors}} = \underbrace{\,5\cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5\vphantom{{}_{\scriptscriptstyle 1}}\,}_{3\ {\rm times}\ 2\ {\rm factors}} = 5^{2 \cdot 3} = 5^6\mbox{}</math>}}

	and		and

-	{{Fristående formel\|\|<math> (5^3)^2 = 5^3\cdot5^3= \underbrace{\,5\cdot 5 \cdot 5\vphantom{{}_{\scriptscriptstyle 1}}\,}_{3\ {\rm st}} \cdot \underbrace{\,5\cdot 5 \cdot 5\vphantom{{}_{\scriptscriptstyle 1}}\,}_{3\ {\rm st}} = \underbrace{\,5\cdot 5 \cdot 5\,\cdot\,5\cdot 5 \cdot 5\vphantom{{}_{\scriptscriptstyle 1}}\,}_{2\ {\rm ~~gånger~~}\ 3\ {\rm st}}=5^{3\cdot2}=5^6\mbox{.}</math>}}	+	{{Fristående formel\|\|<math> (5^3)^2 = 5^3\cdot5^3= \underbrace{\,5\cdot 5 \cdot 5\vphantom{{}_{\scriptscriptstyle 1}}\,}_{3\ {\rm factors}} \cdot \underbrace{\,5\cdot 5 \cdot 5\vphantom{{}_{\scriptscriptstyle 1}}\,}_{3\ {\rm factors}} = \underbrace{\,5\cdot 5 \cdot 5\,\cdot\,5\cdot 5 \cdot 5\vphantom{{}_{\scriptscriptstyle 1}}\,}_{2\ {\rm times}\ 3\ {\rm factors}}=5^{3\cdot2}=5^6\mbox{.}</math>}}


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-	The only way for the rules of ~~powers~~ to agree is to make the	+	The only way for the rules of exponents to agree is to make the
	following but natural definition that for all non zero ''a'' one has that		following but natural definition that for all non zero ''a'' one has that

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	If the base of a power is <math>-1</math> then the expression will simplify to either <math>-1</math> or <math>+1</math> depending on the value of the exponent		If the base of a power is <math>-1</math> then the expression will simplify to either <math>-1</math> or <math>+1</math> depending on the value of the exponent

-	{{Fristående formel\|\|<math>\eqalign{(-1)^1 &= -1\cr (-1)^2 &= (-1)\cdot(-1) = +1\cr (-1)^3 &= (-1)\cdot(-1)^2 = (-1)\cdot 1 = -1\cr (-1)^4 &= (-1)\cdot(-1)^3 = (-1)\cdot (-1) = +1\cr \quad\hbox{~~osv~~.}}</math>}}	+	{{Fristående formel\|\|<math>\eqalign{(-1)^1 &= -1\cr (-1)^2 &= (-1)\cdot(-1) = +1\cr (-1)^3 &= (-1)\cdot(-1)^2 = (-1)\cdot 1 = -1\cr (-1)^4 &= (-1)\cdot(-1)^3 = (-1)\cdot (-1) = +1\cr \quad\hbox{etc.}}</math>}}

	The rule is that <math>(-1)^n </math> is equal to<math>-1</math>		The rule is that <math>(-1)^n </math> is equal to<math>-1</math>
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	</div>		</div>

-	If a power has a positive exponent, it will get larger the larger the base becomes. The opposite applies if the exponent is negative: that is the power decreases as the base gets larger.	+	If a power has a positive exponent, it will get larger the larger the base becomes. The opposite applies if the exponent is negative: that is, the power decreases as the base gets larger.

	<div class="exempel">		<div class="exempel">

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-	{{~~Vald flik~~\|[[1.3 Powers\|Theory]]}}	+	{{Selected tab\|[[1.3 Powers\|Theory]]}}
	{{Not selected tab\|[[1.3 Exercises\|Exercises]]}}		{{Not selected tab\|[[1.3 Exercises\|Exercises]]}}
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	{{Vald flik\|[[1.3 Powers\|Theory]]}}		{{Vald flik\|[[1.3 Powers\|Theory]]}}
-	{{~~Ej vald flik~~\|[[1.3 Exercises\|Exercises]]}}	+	{{Not selected tab\|[[1.3 Exercises\|Exercises]]}}
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-	{{Vald flik\|[[1.3 ~~Potenser~~\|Theory]]}}	+	{{Vald flik\|[[1.3 Powers\|Theory]]}}
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