1.3 Powers

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(Ny sida: __NOTOC__ {{Info| '''Innehåll:''' * Positiv heltalsexponent * Negativ heltalsexponent * Rationell exponent * Potenslagar }} {{Info| '''Lärandemål:''' Efter detta avsnitt ska du ha lär...)
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__NOTOC__
__NOTOC__
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{| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"
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| style="border-bottom:1px solid #797979" width="5px" |  
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{{Selected tab|[[1.3 Powers|Theory]]}}
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{{Not selected tab|[[1.3 Exercises|Exercises]]}}
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| style="border-bottom:1px solid #797979" width="100%"|  
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|}
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{{Info|
{{Info|
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'''Innehåll:'''
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'''Content: '''
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* Positiv heltalsexponent
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* Positive integer exponent
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* Negativ heltalsexponent
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* Negative integer exponent
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* Rationell exponent
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* Rational exponents
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* Potenslagar
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* Laws of exponents
}}
}}
{{Info|
{{Info|
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'''Lärandemål:'''
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'''Learning outcomes:'''
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Efter detta avsnitt ska du ha lärt dig att:
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After this section you will have learned to:
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*Känna till begreppen bas och exponent.
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* Recognise the concepts of base and exponent.
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*Beräkna uttryck med heltalsexponent.
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*Calculate integer power expressions.
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*Hantera potenslagarna i förenkling av potensuttryck.
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*Use the laws of exponents to simplify expressions containing powers.
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*Veta när potenslagarna är giltiga (positiv bas).
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* Know when the laws of exponents are applicable (positive basis).
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*Avgöra vilket av två potensuttryck som är störst baserat på jämförelse av bas/exponent.
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*Determine which of two powers is the larger based on a comparison of the base / exponent.
}}
}}
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== Heltalspotenser ==
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== Integer exponents ==
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Vi använder multiplikationssymbolen som ett kortare skrivsätt
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We use the multiplication symbol as a shorthand for repeated addition of the same number. For example:
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för upprepad addition av samma tal, t.ex.
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{{Fristående formel||<math>4 + 4 + 4 + 4 + 4 = 4 \cdot 5\mbox{.}</math>}}
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{{Displayed math||<math>4 + 4 + 4 + 4 + 4 = 4 \cdot 5\mbox{.}</math>}}
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På ett liknande sätt används potenser som ett kortare skrivsätt för upprepad multiplikation av samma tal:
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In a similar way we use exponentials as a short-hand for repeated multiplication
 +
of the same number:
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{{Fristående formel||<math> 4 \cdot 4 \cdot 4 \cdot 4 \cdot 4 = 4^5\mbox{.}</math>}}
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{{Displayed math||<math> 4 \cdot 4 \cdot 4 \cdot 4 \cdot 4 = 4^5\mbox{.}</math>}}
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Siffran 4 kallas för potensens ''bas'' och siffran 5 dess ''exponent''.
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The 4 is called the base of the power and the 5 is its exponent.
<div class="exempel">
<div class="exempel">
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'''Exempel 1'''
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'''Example 1 '''
<ol type="a">
<ol type="a">
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= 10 \cdot 10 \cdot 10 \cdot 10 \cdot 10 = 100 000</math></li>
= 10 \cdot 10 \cdot 10 \cdot 10 \cdot 10 = 100 000</math></li>
<li><math>0{,}1^3
<li><math>0{,}1^3
-
= 0{,}1 \cdot 0{,}1 \cdot 0{,}1 = 0{,}001</math></li>
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= 0\text{.}1 \cdot 0\text{.}1 \cdot 0\text{.}1 = 0\text{.}001</math></li>
<li><math>(-2)^4
<li><math>(-2)^4
-
= (-2) \cdot (-2) \cdot (-2) \cdot (-2)= 16</math>, men <math> -2^4
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= (-2) \cdot (-2) \cdot (-2) \cdot (-2)= 16</math>, but <math> -2^4
= -(2^4) = - (2 \cdot 2 \cdot 2 \cdot 2) = -16</math></li>
= -(2^4) = - (2 \cdot 2 \cdot 2 \cdot 2) = -16</math></li>
-
<li><math> 2\cdot 3^2 = 2 \cdot 3 \cdot 3 = 18</math>, men <math>
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<li><math> 2\cdot 3^2 = 2 \cdot 3 \cdot 3 = 18</math>, but <math>
(2\cdot3)^2 = 6^2 = 36</math></li>
(2\cdot3)^2 = 6^2 = 36</math></li>
</ol>
</ol>
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<div class="exempel">
<div class="exempel">
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'''Exempel 2'''
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'''Example 2'''
<ol type="a">
<ol type="a">
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</div>
</div>
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Det sista exemplet kan generaliseras till två användbara räkneregler för potenser:
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The last example can be generalised to two useful rules when calculating powers:
<div class="regel">
<div class="regel">
-
{{Fristående formel||<math>\left(\displaystyle\frac{a}{b}\right)^m = \displaystyle\frac{a^m}{b^m} \quad \mbox{och}\quad (ab)^m = a^m b^m</math>}}
+
{{Displayed math||<math>\left(\displaystyle\frac{a}{b}\right)^m = \displaystyle\frac{a^m}{b^m} \quad \mbox{and}\quad (ab)^m = a^m b^m\,\mbox{.}</math>}}
</div>
</div>
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== Potenslagar ==
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== Laws of exponents ==
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Med definitionen av potens följer ytterligare några räkneregler som förenklar beräkningar med potenser inblandade. Man ser t.ex. att
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There are a few more rules that lead on from the definition of power which are useful when performing 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>}}
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{{Displayed math||<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\text{,}</math>}}
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vilket generellt kan skrivas
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which generally can be expressed as
<div class="regel">
<div class="regel">
-
{{Fristående formel||<math>a^m \cdot a^n = a^{m+n}\mbox{.}</math>}}
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{{Displayed math||<math>a^m \cdot a^n = a^{m+n}\mbox{.}</math>}}
</div>
</div>
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Vid division av potenser kan också beräkningarna förenklas om potenserna har samma bas
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There is also a useful simplification rule for the division of powers that have the same base.
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{{Fristående formel||<math>\frac{2^7}{2^3}=\displaystyle\frac{ 2\cdot 2\cdot 2\cdot 2\cdot \not{2}\cdot \not{2}\cdot \not{2} }{ \not{2}\cdot \not{2}\cdot \not{2}} = 2^{7-3}=2^4\mbox{.}</math>}}
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{{Displayed math||<math>\frac{2^7}{2^3}=\displaystyle\frac{ 2\cdot 2\cdot 2\cdot 2\cdot \not{2}\cdot \not{2}\cdot \not{2} }{ \not{2}\cdot \not{2}\cdot \not{2}} = 2^{7-3}=2^4\mbox{.}</math>}}
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Den allmänna regeln blir
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The general rule is
<div class="regel">
<div class="regel">
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{{Fristående formel||<math>\displaystyle\frac{a^m}{a^n}= a^{m-n}\mbox{.}</math>}}
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{{Displayed math||<math>\displaystyle\frac{a^m}{a^n}= a^{m-n}\mbox{.}</math>}}
</div>
</div>
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När man råkar ut för en potens av en potens finns ytterligare
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For the case when the base itself is a power there is another useful rule. We see that
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en användbar räkneregel. Vi ser att
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-
{{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>}}
+
{{Displayed math||<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>}}
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och
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and
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{{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^{2\cdot3}=5^6\mbox{.}</math>}}
+
{{Displayed math||<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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Allmänt kan detta skrivas
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Generally, this can be written
<div class="regel">
<div class="regel">
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{{Fristående formel||<math>(a^m)^n = a^{m \cdot n}\mbox{.} </math>}}
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{{Displayed math||<math>(a^m)^n = a^{m \cdot n}\mbox{.} </math>}}
</div>
</div>
<div class="exempel">
<div class="exempel">
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'''Exempel 3'''
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''' Example 3'''
<ol type="a">
<ol type="a">
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<div class="exempel">
<div class="exempel">
-
'''Exempel 4'''
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'''Example 4'''
<ol type="a">
<ol type="a">
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Om ett bråk har samma potensuttryck i både täljare och nämnare så inträffar följande:
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If a fraction has the same expression for the exponent in both the numerator and the denominator we can simplify in two ways:
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{{Fristående formel||<math>\frac{5^3}{5^3} = 5^{3-3} = 5^0\quad\text{samtidigt som}\quad \frac{5^3}{5^3} = \frac{ 5 \cdot 5 \cdot 5 }{ 5 \cdot 5 \cdot 5 } = \frac{125}{125} = 1\mbox{.}</math>}}
+
{{Displayed math||<math>\frac{5^3}{5^3} = 5^{3-3} = 5^0\quad\text{as well as}\quad \frac{5^3}{5^3} = \frac{ 5 \cdot 5 \cdot 5 }{ 5 \cdot 5 \cdot 5 } = \frac{125}{125} = 1\mbox{.}</math>}}
 +
 
 +
The only way for the rules of exponents to agree is to make the
 +
following natural definition. For all non zero ''a'' we have
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För att räknereglerna för potenser ska stämma gör man alltså den naturliga definitionen att för alla ''a'' som inte är 0 gäller att
 
<div class="regel">
<div class="regel">
-
{{Fristående formel||<math> a^0 = 1\mbox{.} </math>}}
+
{{Displayed math||<math> a^0 = 1\mbox{.} </math>}}
</div>
</div>
-
Vi kan också råka ut för att exponenten i nämnaren är större än den i täljaren. Vi får t.ex.
+
We can also run into examples where the exponent in the denominator is greater than that in the numerator. For example we have
-
{{Fristående formel||<math>\frac{3^4}{3^6} = 3^{4-6} = 3^{-2}\quad\text{och}\quad \frac{3^4}{3^6} = \frac{\not{3} \cdot \not{3} \cdot \not{3} \cdot \not{3} }{ \not{3} \cdot \not{3} \cdot \not{3} \cdot \not{3} \cdot 3 \cdot 3} = \frac{1}{3 \cdot 3} = \frac{1}{3^2}\mbox{.}</math>}}
+
{{Displayed math||<math>\frac{3^4}{3^6} = 3^{4-6} = 3^{-2}\quad\text{and}\quad \frac{3^4}{3^6} = \frac{\not{3} \cdot \not{3} \cdot \not{3} \cdot \not{3} }{ \not{3} \cdot \not{3} \cdot \not{3} \cdot \not{3} \cdot 3 \cdot 3} = \frac{1}{3 \cdot 3} = \frac{1}{3^2}\mbox{.}</math>}}
-
Vi ser här att enligt våra räkneregler måste den negativa exponenten betyda att
+
It is therefore necessary that we assume the negative sign of the exponent implies that
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{{Fristående formel||<math>3^{-2} = \frac{1}{3^2}\mbox{.}</math>}}
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{{Displayed math||<math>3^{-2} = \frac{1}{3^2}\mbox{.}</math>}}
-
Den allmänna definitionen av negativa exponenter är att, för alla tal ''a'' som inte är 0 gäller att
 
 +
We therefore note that the general definition for negative exponents is that for all non zero numbers ''a'', we have
 +
of as follows
<div class="regel">
<div class="regel">
-
{{Fristående formel||<math>a^{-n} = \frac{1}{a^n}\mbox{.}</math>}}
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{{Displayed math||<math>a^{-n} = \frac{1}{a^n}\mbox{.}</math>}}
</div>
</div>
<div class="exempel">
<div class="exempel">
-
'''Exempel 5'''
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''' Example 5'''
<ol type="a">
<ol type="a">
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</div>
</div>
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Om basen i ett potensuttryck är <math>-1</math> så blir uttrycket alternerande <math>-1</math> eller <math>+1</math> beroende på exponentens värde
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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
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{{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{o.s.v.}}</math>}}
+
{{Displayed math||<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>}}
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Regeln är att <math>(-1)^n</math> är lika med <math>-1</math> om <math>n</math> är udda och lika med <math>+1</math> om <math>n</math> är jämn.
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The rule is that <math>(-1)^n </math> is equal to <math>-1</math>
 +
if <math>n</math> is odd and equal to <math>+1</math> if <math>n</math> is even .
<div class="exempel">
<div class="exempel">
-
'''Exempel 6'''
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''' Example 6'''
<ol type="a">
<ol type="a">
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<li><math>(-1)^{56} = 1\quad</math> eftersom <math>56</math> är ett jämnt tal</li>
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<li><math>(-1)^{56} = 1\quad</math> as <math>56</math> is an even number </li>
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<li><math>\frac{1}{(-1)^{11}} = \frac{1}{-1} = -1\quad</math> eftersom 11 är ett udda tal</li>
+
<li><math>\frac{1}{(-1)^{11}} = \frac{1}{-1} = -1\quad</math> because 11 is an odd number </li>
<li><math>\frac{(-2)^{127}}{2^{130}} = \frac{(-1 \cdot 2)^{127}}{2^{130}}
<li><math>\frac{(-2)^{127}}{2^{130}} = \frac{(-1 \cdot 2)^{127}}{2^{130}}
= \frac{(-1)^{127} \cdot 2^{127}}{2^{130}}
= \frac{(-1)^{127} \cdot 2^{127}}{2^{130}}
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== Byte av bas ==
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==Changing the base ==
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Man bör vara uppmärksam på att vid förenkling av uttryck om möjligt försöka skriva ihop potenser genom att välja samma bas. Det handlar ofta om att välja 2, 3 eller 5 som bas och därför bör man lära sig att känna igen potenser av dessa tal, exempelvis
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A point to observe is that when simplifying expressions one should try if possible to combine powers by choosing the same base. This often involves selecting 2, 3 or 5 as a base. It is therefore a good idea to learn to recognise the powers of these numbers, such as:
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{{Fristående formel||<math>4=2^2,\;\; 8=2^3,\;\; 16=2^4,\;\; 32=2^5,\;\; 64=2^6,\;\; 128=2^7,\;\ldots</math>}}
+
{{Displayed math||<math>4=2^2,\;\; 8=2^3,\;\; 16=2^4,\;\; 32=2^5,\;\; 64=2^6,\;\; 128=2^7,\;\ldots</math>}}
-
{{Fristående formel||<math>9=3^2,\;\; 27=3^3,\;\; 81=3^4,\;\; 243=3^5,\;\ldots</math>}}
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{{Displayed math||<math>9=3^2,\;\; 27=3^3,\;\; 81=3^4,\;\; 243=3^5,\;\ldots</math>}}
-
{{Fristående formel||<math>25=5^2,\;\; 125=5^3,\;\; 625=5^4,\;\ldots</math>}}
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{{Displayed math||<math>25=5^2,\;\; 125=5^3,\;\; 625=5^4,\;\ldots</math>}}
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Men även
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Similarly, one should become familiar with
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{{Fristående formel||<math>\frac{1}{4}=\frac{1}{2^2} = 2^{-2},\;\; \frac{1}{8}=\frac{1}{2^3}=2^{-3},\;\; \frac{1}{16}=\frac{1}{2^4}=2^{-4},\;\ldots</math>}}
+
{{Displayed math||<math>\frac{1}{4}=\frac{1}{2^2} = 2^{-2},\;\; \frac{1}{8}=\frac{1}{2^3}=2^{-3},\;\; \frac{1}{16}=\frac{1}{2^4}=2^{-4},\;\ldots</math>}}
-
{{Fristående formel||<math>\frac{1}{9}=\frac{1}{3^2}=3^{-2},\;\; \frac{1}{27}=\frac{1}{3^3}=3^{-3},\;\ldots</math>}}
+
{{Displayed math||<math>\frac{1}{9}=\frac{1}{3^2}=3^{-2},\;\; \frac{1}{27}=\frac{1}{3^3}=3^{-3},\;\ldots</math>}}
-
{{Fristående formel||<math>\frac{1}{25}=\frac{1}{5^2}=5^{-2},\;\; \frac{1}{125}=\frac{1}{5^3}=5^{-3},\;\ldots</math>}}
+
{{Displayed math||<math>\frac{1}{25}=\frac{1}{5^2}=5^{-2},\;\; \frac{1}{125}=\frac{1}{5^3}=5^{-3},\;\ldots</math>}}
-
osv.
+
and so on.
<div class="exempel">
<div class="exempel">
-
'''Exempel 7'''
+
''' Example 7'''
<ol type="a">
<ol type="a">
-
<li>Skriv <math>\ 8^3 \cdot 4^{-2} \cdot 16\ </math> som en potens med basen 2.
+
<li> Write <math>\ 8^3 \cdot 4^{-2} \cdot 16\ </math> as a power with base 2
<br/>
<br/>
<br/>
<br/>
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:<math>\qquad\quad{}= 2^9 \cdot 2^{-4} \cdot 2^4 = 2^{9-4+4} =2^9</math></li>
:<math>\qquad\quad{}= 2^9 \cdot 2^{-4} \cdot 2^4 = 2^{9-4+4} =2^9</math></li>
-
<li>Skriv <math>\ \frac{27^2 \cdot (1/9)^{-2}}{81^2}\ </math> som en potens av basen 3.
+
<li> Write <math>\ \frac{27^2 \cdot (1/9)^{-2}}{81^2}\ </math> as a power with base 3.
<br/>
<br/>
<br/>
<br/>
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:<math>\qquad\quad{} = \frac{3^{3 \cdot 2} \cdot 3^{(-2) \cdot (-2)}}{3^{4 \cdot 2}} = \frac{3^6\cdot 3^4}{3^8} = \frac{3^{6 + 4}}{3^8}= \frac{3^{10}}{3^8} = 3^{10-8}= 3^2</math></li>
:<math>\qquad\quad{} = \frac{3^{3 \cdot 2} \cdot 3^{(-2) \cdot (-2)}}{3^{4 \cdot 2}} = \frac{3^6\cdot 3^4}{3^8} = \frac{3^{6 + 4}}{3^8}= \frac{3^{10}}{3^8} = 3^{10-8}= 3^2</math></li>
-
<li>Skriv <math>\frac{81 \cdot 32^2 \cdot (2/3)^2}{2^5+2^4}</math> så enkelt som möjligt.
+
<li> Write <math>\frac{81 \cdot 32^2 \cdot (2/3)^2}{2^5+2^4}</math> in as simple a form as possible.
<br/>
<br/>
<br/>
<br/>
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-
== Rationell exponent ==
+
== Rational exponents ==
-
Vad händer om ett tal höjs upp till en rationell exponent? Gäller fortfarande de definitioner och räkneregler vi har använt oss av ovan?
+
What happens if a number is raised to a rational exponent? Do the definitions and the rules we have used in the above calculations still hold?
-
Eftersom exempelvis
+
For instance we note that
-
{{Fristående formel||<math>2^{1/2} \cdot 2^{1/2} = 2^{1/2 + 1/2} = 2^1 = 2</math>}}
+
{{Displayed math||<math>2^{1/2} \cdot 2^{1/2} = 2^{1/2 + 1/2} = 2^1 = 2</math>}}
-
så måste <math> 2^{1/2} </math> vara samma sak som <math>\sqrt{2}</math> i och med att <math>\sqrt2</math> definieras som det tal som uppfyller <math>\sqrt2\cdot\sqrt2 = 2</math>&nbsp;.
+
so <math> 2^{1/2} </math> must be the same as <math>\sqrt{2}</math>. This is because <math>\sqrt2</math> is defined as the number which satisfies <math>\sqrt2\cdot\sqrt2 = 2</math>&nbsp;.
-
Allmänt kan vi göra definitionen
+
Generally, we define
<div class="regel">
<div class="regel">
-
{{Fristående formel||<math>a^{1/2} = \sqrt{a}\mbox{.}</math>}}
+
{{Displayed math||<math>a^{1/2} = \sqrt{a}\mbox{.}</math>}}
</div>
</div>
-
Vi måste då förutsätta att <math>a\ge 0</math>, eftersom inget reellt tal multiplicerat med sig själv kan ge ett negativt tal.
+
We must assume that <math>a\ge 0</math> since no real number multiplied by itself can give a negative number.
-
Man ser också att exempelvis
+
We also see that, for example,
-
{{Fristående formel||<math>5^{1/3} \cdot 5^{1/3} \cdot 5^{1/3} = 5^{1/3 + 1/3 +1/3} = 5^1 = 5</math>}}
+
{{Displayed math||<math>5^{1/3} \cdot 5^{1/3} \cdot 5^{1/3} = 5^{1/3 + 1/3 +1/3} = 5^1 = 5</math>}}
-
som innebär att <math>\,5^{1/3} = \sqrt[\scriptstyle3]{5}\mbox{,}\,</math> vilket kan generaliseras till att
+
which implies that <math>\,5^{1/3} = \sqrt[\scriptstyle3]{5}\mbox{,}\,</math>. This can be generalised to
<div class="regel">
<div class="regel">
-
{{Fristående formel||<math>a^{1/n} = \sqrt[\scriptstyle n]{a}\mbox{.}</math>}}
+
{{Displayed math||<math>a^{1/n} = \sqrt[\scriptstyle n]{a}\mbox{.}</math>}}
</div>
</div>
-
Genom att kombinera denna definition med en av de tidigare potenslagarna <math>((a^m)^n=a^{m\cdot n})</math> får vi att, för alla <math>a\ge0</math> gäller att
+
By combining this definition with one of our previous laws for exponents, namely <math>((a^m)^n=a^{m\cdot n})</math>, we have that for all <math>a\ge0</math>, the following holds:
<div class="regel">
<div class="regel">
-
{{Fristående formel||<math>a^{m/n} = (a^m)^{1/n} = \sqrt[\scriptstyle n]{a^m}</math>}}
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{{Displayed math||<math>a^{m/n} = (a^m)^{1/n} = \sqrt[\scriptstyle n]{a^m}</math>}}
-
eller
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or alternatively
-
{{Fristående formel||<math>a^{m/n} = (a^{1/n})^m = (\sqrt[\scriptstyle n]{a}\,)^m\mbox{.} </math>}}
+
{{Displayed math||<math>a^{m/n} = (a^{1/n})^m = (\sqrt[\scriptstyle n]{a}\,)^m\mbox{.} </math>}}
</div>
</div>
<div class="exempel">
<div class="exempel">
-
'''Exempel 8'''
+
''' Example 8'''
<ol type="a">
<ol type="a">
<li><math>27^{1/3} = \sqrt[\scriptstyle 3]{27}
<li><math>27^{1/3} = \sqrt[\scriptstyle 3]{27}
-
= 3\quad</math> eftersom <math>3 \cdot 3 \cdot 3 =27</math></li>
+
= 3\quad</math> as <math>3 \cdot 3 \cdot 3 =27</math></li>
<li><math>1000^{-1/3} = \frac{1}{1000^{1/3}}
<li><math>1000^{-1/3} = \frac{1}{1000^{1/3}}
= \frac{1}{(10^3)^{1/3}}
= \frac{1}{(10^3)^{1/3}}
Line 302: Line 312:
-
== Jämförelse av potenser ==
+
==Comparison of powers ==
-
Om man utan tillgång till miniräknare vill jämföra storleken av potenser, kan man i vissa fall avgöra detta genom att jämföra basen eller exponenten.
+
If we do not have access to calculators and wish to compare the size of powers, one can sometimes achieve this by comparing bases or exponents.
-
Om basen i en potens är större är än <math>1</math> så blir potensen större ju större exponenten är. Är däremot basen mellan <math>0</math> och <math>1</math> så blir potensen mindre istället när exponenten växer.
+
If the base of a power is greater than <math>1</math> then the power increases as the exponent increases. On the other hand, if the base lies between <math>0</math> and <math>1</math> then the power decreases as the exponent grows.
<div class="exempel">
<div class="exempel">
-
'''Exempel 9'''
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''' Example 9'''
<ol type="a">
<ol type="a">
-
<li><math>\quad 3^{5/6} > 3^{3/4}\quad</math> eftersom basen <math>3</math> är större än <math>1</math> och den första exponenten <math>5/6</math> är större än den andra exponenten <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> eftersom basen är större än <math>1</math> och exponenterna uppfyller <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> eftersom basen <math> 0{,}3</math> är mellan <math>0</math> och <math>1</math> och <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>
-
Har en potens en positiv exponent så blir potensen större ju större basen är. Det motsatta gäller om exponenten är negativ: då blir potensen mindre när basen blir större.
+
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">
-
'''Exempel 10'''
+
''' Example 10'''
<ol type="a">
<ol type="a">
-
<li><math>\quad 5^{3/2} > 4^{3/2}\quad</math> eftersom basen <math>5</math> är större än basen <math>4</math> och båda potenserna har samma positiva exponenten <math>3/2</math>.</li>
+
<li><math>\quad 5^{3/2} > 4^{3/2}\quad</math> as the base <math>5</math> is larger than the base <math>4</math> and both powers have the same positive exponent <math>3/2</math>.</li>
-
<li><math> \quad 2^{-5/3} > 3^{-5/3}\quad</math> eftersom baserna uppfyller <math>2<3</math> och potenserna har den negativa exponenten <math>-5/3</math>.</li>
+
<li><math> \quad 2^{-5/3} > 3^{-5/3}\quad</math> as the bases satisfy <math>2<3</math> and the powers have a negative exponent <math>-5/3</math>.</li>
</ol>
</ol>
</div>
</div>
-
Ibland krävs det en omskrivning av potenserna för att kunna avgöra storleksförhållandet. Vill man t.ex. jämföra <math>125^2</math> med <math>36^3</math> kan man göra omskrivningarna
+
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
-
{{Fristående formel||<math>
+
{{Displayed math||<math>
-
125^2 = (5^3)^2 = 5^6\quad \text{och}\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
</math>}}
</math>}}
-
varefter man kan konstatera att <math>36^3 > 125^2</math>.
+
after which we see that <math>36^3 > 125^2</math>.
<div class="exempel">
<div class="exempel">
-
'''Exempel 11'''
+
''' Example 11'''
-
Avgör vilket tal som är störst av
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Determine which of the following pairs of numbers is the greater:
<ol type="a">
<ol type="a">
-
<li><math> 25^{1/3} </math>&nbsp; och &nbsp;<math> 5^{3/4} </math>
+
<li><math> 25^{1/3} </math>&nbsp; and &nbsp;<math> 5^{3/4} </math>.
<br>
<br>
<br>
<br>
-
Basen 25 kan skrivas om i termer av den andra basen <math>5</math> genom att <math>25= 5\cdot 5= 5^2</math>. Därför är
+
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
-
{{Fristående formel||<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>}}
-
och då ser vi att
+
hence we see that
-
{{Fristående formel||<math>5^{3/4} > 25^{1/3} </math>}}
+
{{Displayed math||<math>5^{3/4} > 25^{1/3} </math>}}
-
eftersom <math>\frac{3}{4} > \frac{2}{3}</math> och basen <math>5</math> är större än <math>1</math>.</li>
+
since <math>\frac{3}{4} > \frac{2}{3}</math> and the base <math>5</math> is larger than <math>1</math>.</li>
-
<li><math>(\sqrt{8}\,)^5 </math>&nbsp; och <math>128</math>
+
<li><math>(\sqrt{8}\,)^5 </math>&nbsp; and <math>128</math>.
<br>
<br>
<br>
<br>
-
Både <math>8</math> och <math>128</math> kan skrivas som potenser av <math>2</math>
+
Both <math>8</math> and <math>128</math> can be written as powers of <math>2</math>
-
{{Fristående formel||<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>}}
-
Detta betyder att
+
This gives
-
{{Fristående formel||<math>\begin{align*}
+
{{Displayed math||<math>\begin{align*}
(\sqrt{8}\,)^5 &= (8^{1/2})^5 = (8)^{5/2} = (2^3)^{5/2}
(\sqrt{8}\,)^5 &= (8^{1/2})^5 = (8)^{5/2} = (2^3)^{5/2}
= 2^{3\cdot\frac{5}{2}}= 2^{15/2}\\
= 2^{3\cdot\frac{5}{2}}= 2^{15/2}\\
Line 370: Line 380:
\end{align*}</math>}}
\end{align*}</math>}}
-
och därför är
+
and thus
-
{{Fristående formel||<math>(\sqrt{8}\,)^5 > 128 </math>}}
+
{{Displayed math||<math>(\sqrt{8}\,)^5 > 128 </math>}}
-
i och med att <math>\frac{15}{2} > \frac{14}{2}</math> och basen <math>2</math> är större än <math>1</math>.</li>
+
because <math>\frac{15}{2} > \frac{14}{2}</math> and the base <math>2</math> is greater than <math>1</math>.</li>
-
<li><math> (8^2)^{1/5} </math> och <math> (\sqrt{27}\,)^{4/5}</math>
+
<li><math> (8^2)^{1/5} </math> and <math> (\sqrt{27}\,)^{4/5}</math>.
<br>
<br>
<br>
<br>
-
Eftersom <math>8=2^3</math> och <math>27=3^3</math> så kan ett första steg vara att förenkla och skriva talen som potenser av <math>2</math> respektive <math>3</math>,
+
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,
-
{{Fristående formel||<math>\begin{align*}
+
{{Displayed math||<math>\begin{align*}
(8^2)^{1/5} &= (8)^{2/5} = (2^3)^{2/5} = 2^{3\cdot \frac{2}{5}}
(8^2)^{1/5} &= (8)^{2/5} = (2^3)^{2/5} = 2^{3\cdot \frac{2}{5}}
= 2^{6/5}\mbox{,}\\
= 2^{6/5}\mbox{,}\\
Line 390: Line 400:
\end{align*}</math>}}
\end{align*}</math>}}
-
Nu ser vi att
+
Now we see that
-
{{Fristående formel||<math>(\sqrt{27}\,)^{4/5} > (8^2)^{1/5} </math>}}
+
{{Displayed math||<math>(\sqrt{27}\,)^{4/5} > (8^2)^{1/5} </math>}}
-
eftersom <math> 3>2</math> och exponenten <math>\frac{6}{5}</math> är positiv.
+
because <math> 3>2</math> and exponent <math>\frac{6}{5}</math> is positive.
-
<li><math> 3^{1/3} </math>&nbsp; och &nbsp;<math> 2^{1/2}</math>
+
<li><math> 3^{1/3} </math>&nbsp; and &nbsp;<math> 2^{1/2}</math>
<br>
<br>
<br>
<br>
-
Vi skriver exponenterna med gemensam nämnare
+
We rewrite the exponents due to them having a common denominator
-
{{Fristående formel||<math>\frac{1}{3} = \frac{2}{6} \quad</math> och <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>.}}
-
Då har vi att
+
This gives
-
{{Fristående formel||<math>\begin{align*}
+
{{Displayed math||<math>\begin{align*}
3^{1/3} &= 3^{2/6} = (3^2)^{1/6} = 9^{1/6}\\
3^{1/3} &= 3^{2/6} = (3^2)^{1/6} = 9^{1/6}\\
2^{1/2} &= 2^{3/6} = (2^3)^{1/6} = 8^{1/6}
2^{1/2} &= 2^{3/6} = (2^3)^{1/6} = 8^{1/6}
\end{align*}</math>}}
\end{align*}</math>}}
-
och vi ser att
+
and we see that
-
{{Fristående formel||<math> 3^{1/3} > 2^{1/2} </math>}}
+
{{Displayed math||<math> 3^{1/3} > 2^{1/2} </math>}}
-
eftersom <math> 9>8</math> och exponenten <math>1/6</math> är positiv.</li>
+
because <math> 9>8</math> and the exponent <math>1/6</math> is positive.</li>
</ol>
</ol>
</div>
</div>
-
[[1.3 Övningar|Övningar]]
+
[[1.3 Exercises|Exercises]]
-
<div class="inforuta">
+
<div class="inforuta" style="width:580px;">
-
'''Råd för inläsning'''
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'''Study advice'''
-
'''Grund- och slutprov'''
+
'''Basic and final tests'''
-
Efter att du har läst texten och arbetat med övningarna ska du göra grund- och slutprovet för att bli godkänd på detta avsnitt. Du hittar länken till proven i din student lounge.
+
After you have read the text and worked through the exercises, you should do the basic and final tests to pass this section. You can find the link to the tests in your student lounge.
-
'''Tänk på att:'''
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'''Keep in mind that...'''
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Ett tal upphöjt till 0 är 1, om talet (basen) är skild från 0.
+
The number raised to the power 0 is always 1 as long as the number (the base) is not 0.
-
'''Lästips'''
+
'''Reviews'''
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för dig som vill fördjupa dig ytterligare eller behöver en längre förklaring
+
For those of you who want to deepen your studies or need more detailed explanations consider the following references
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[http://en.wikipedia.org/wiki/Exponent Läs mer om potenser på engelska Wikipedia]
+
[http://en.wikipedia.org/wiki/Exponent Learn more about powers in the English Wikipedi]
-
[http://primes.utm.edu/ Vilket är det största primtalet? Läs mer på The Prime Pages]
+
[http://primes.utm.edu/ What is the greatest prime number? Read more at The Prime Page]
-
'''Länktips'''
+
'''Useful web sites'''
-
[http://www.ltcconline.net/greenl/java/BasicAlgebra/ExponentRules/ExponentRules.html Här kan du träna på potenslagarna]
+
[http://www.ltcconline.net/greenl/java/BasicAlgebra/ExponentRules/ExponentRules.html Here you can practise the laws of exponents]
</div>
</div>

Current revision

       Theory          Exercises      

Content:

  • Positive integer exponent
  • Negative integer exponent
  • Rational exponents
  • Laws of exponents

Learning outcomes:

After this section you will have learned to:

  • Recognise the concepts of base and exponent.
  • Calculate integer power expressions.
  • Use the laws of exponents to simplify expressions containing powers.
  • Know when the laws of exponents are applicable (positive basis).
  • Determine which of two powers is the larger based on a comparison of the base / exponent.

Integer exponents

We use the multiplication symbol as a shorthand for repeated addition of the same number. For example:

\displaystyle 4 + 4 + 4 + 4 + 4 = 4 \cdot 5\mbox{.}

In a similar way we use exponentials as a short-hand for repeated multiplication of the same number:

\displaystyle 4 \cdot 4 \cdot 4 \cdot 4 \cdot 4 = 4^5\mbox{.}

The 4 is called the base of the power and the 5 is its exponent.

Example 1

  1. \displaystyle 5^3 = 5 \cdot 5 \cdot 5 = 125
  2. \displaystyle 10^5 = 10 \cdot 10 \cdot 10 \cdot 10 \cdot 10 = 100 000
  3. \displaystyle 0{,}1^3 = 0\text{.}1 \cdot 0\text{.}1 \cdot 0\text{.}1 = 0\text{.}001
  4. \displaystyle (-2)^4 = (-2) \cdot (-2) \cdot (-2) \cdot (-2)= 16, but \displaystyle -2^4 = -(2^4) = - (2 \cdot 2 \cdot 2 \cdot 2) = -16
  5. \displaystyle 2\cdot 3^2 = 2 \cdot 3 \cdot 3 = 18, but \displaystyle (2\cdot3)^2 = 6^2 = 36

Example 2

  1. \displaystyle \left(\displaystyle\frac{2}{3}\right)^3 = \displaystyle\frac{2}{3}\cdot \displaystyle\frac{2}{3} \cdot \displaystyle\frac{2}{3} = \displaystyle\frac{2^3}{3^3} = \displaystyle\frac{8}{27}
  2. \displaystyle (2\cdot 3)^4 = (2\cdot 3)\cdot(2\cdot 3)\cdot(2\cdot 3)\cdot(2\cdot 3)
    \displaystyle \phantom{(2\cdot 3)^4}{} = 2\cdot 2\cdot 2\cdot 2\cdot 3\cdot 3\cdot 3\cdot 3 = 2^4 \cdot 3^4 = 1296

The last example can be generalised to two useful rules when calculating powers:

\displaystyle \left(\displaystyle\frac{a}{b}\right)^m = \displaystyle\frac{a^m}{b^m} \quad \mbox{and}\quad (ab)^m = a^m b^m\,\mbox{.}


Laws of exponents

There are a few more rules that lead on from the definition of power which are useful when performing calculations. You can see for example that

\displaystyle 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\text{,}

which generally can be expressed as

\displaystyle a^m \cdot a^n = a^{m+n}\mbox{.}

There is also a useful simplification rule for the division of powers that have the same base.

\displaystyle \frac{2^7}{2^3}=\displaystyle\frac{ 2\cdot 2\cdot 2\cdot 2\cdot \not{2}\cdot \not{2}\cdot \not{2} }{ \not{2}\cdot \not{2}\cdot \not{2}} = 2^{7-3}=2^4\mbox{.}

The general rule is

\displaystyle \displaystyle\frac{a^m}{a^n}= a^{m-n}\mbox{.}

For the case when the base itself is a power there is another useful rule. We see that

\displaystyle (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{}

and

\displaystyle (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{.}


Generally, this can be written

\displaystyle (a^m)^n = a^{m \cdot n}\mbox{.}

Example 3

  1. \displaystyle 2^9 \cdot 2^{14} = 2^{9+14} = 2^{23}
  2. \displaystyle 5\cdot5^3 = 5^1\cdot5^3 = 5^{1+3} = 5^4
  3. \displaystyle 3^2 \cdot 3^3 \cdot 3^4 = 3^{2+3+4} = 3^9
  4. \displaystyle 10^5 \cdot 1000 = 10^5 \cdot 10^3 = 10^{5+3} = 10^8

Example 4

  1. \displaystyle \frac{3^{100}}{3^{98}} = 3^{100-98} = 3^2
  2. \displaystyle \frac{7^{10}}{7} = \frac{7^{10}}{7^1} = 7^{10-1} = 7^9


If a fraction has the same expression for the exponent in both the numerator and the denominator we can simplify in two ways:

\displaystyle \frac{5^3}{5^3} = 5^{3-3} = 5^0\quad\text{as well as}\quad \frac{5^3}{5^3} = \frac{ 5 \cdot 5 \cdot 5 }{ 5 \cdot 5 \cdot 5 } = \frac{125}{125} = 1\mbox{.}


The only way for the rules of exponents to agree is to make the following natural definition. For all non zero a we have


\displaystyle a^0 = 1\mbox{.}

We can also run into examples where the exponent in the denominator is greater than that in the numerator. For example we have

\displaystyle \frac{3^4}{3^6} = 3^{4-6} = 3^{-2}\quad\text{and}\quad \frac{3^4}{3^6} = \frac{\not{3} \cdot \not{3} \cdot \not{3} \cdot \not{3} }{ \not{3} \cdot \not{3} \cdot \not{3} \cdot \not{3} \cdot 3 \cdot 3} = \frac{1}{3 \cdot 3} = \frac{1}{3^2}\mbox{.}

It is therefore necessary that we assume the negative sign of the exponent implies that

\displaystyle 3^{-2} = \frac{1}{3^2}\mbox{.}


We therefore note that the general definition for negative exponents is that for all non zero numbers a, we have of as follows

\displaystyle a^{-n} = \frac{1}{a^n}\mbox{.}


Example 5

  1. \displaystyle \frac{7^{1293}}{7^{1293}} = 7^{1293 - 1293} = 7^0 = 1
  2. \displaystyle 3^7 \cdot 3^{-9} \cdot 3^4 = 3^{7+(-9)+4} = 3^2
  3. \displaystyle 0{,}001 = \frac{1}{1000} = \frac{1}{10^3} = 10^{-3}
  4. \displaystyle 0{,}008 = \frac{8}{1000} = \frac{1}{125} = \frac{1}{5^3} = 5^{-3}
  5. \displaystyle \left(\frac{2}{3}\right)^{-1} = \frac{1}{\displaystyle\left(\frac{2}{3}\right)^1} = 1\cdot \frac{3}{2} = \frac{3}{2}
  6. \displaystyle \left(\frac{1}{3^2}\right)^{-3} = (3^{-2})^{-3} = 3^{(-2)\cdot(-3)}=3^6
  7. \displaystyle 0.01^5 = (10^{-2})^5 = 10^{-2 \cdot 5} = 10^{-10}

If the base of a power is \displaystyle -1 then the expression will simplify to either \displaystyle -1 or \displaystyle +1 depending on the value of the exponent

\displaystyle \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.}}

The rule is that \displaystyle (-1)^n is equal to \displaystyle -1 if \displaystyle n is odd and equal to \displaystyle +1 if \displaystyle n is even .


Example 6

  1. \displaystyle (-1)^{56} = 1\quad as \displaystyle 56 is an even number
  2. \displaystyle \frac{1}{(-1)^{11}} = \frac{1}{-1} = -1\quad because 11 is an odd number
  3. \displaystyle \frac{(-2)^{127}}{2^{130}} = \frac{(-1 \cdot 2)^{127}}{2^{130}} = \frac{(-1)^{127} \cdot 2^{127}}{2^{130}} = \frac{-1 \cdot 2^{127}}{2^{130}} \displaystyle \phantom{\frac{(-2)^{127}}{2^{130}}}{} = - 2^{127-130} = -2^{-3} = - \frac{1}{2^3} = - \frac{1}{8}


Changing the base

A point to observe is that when simplifying expressions one should try if possible to combine powers by choosing the same base. This often involves selecting 2, 3 or 5 as a base. It is therefore a good idea to learn to recognise the powers of these numbers, such as:

\displaystyle 4=2^2,\;\; 8=2^3,\;\; 16=2^4,\;\; 32=2^5,\;\; 64=2^6,\;\; 128=2^7,\;\ldots
\displaystyle 9=3^2,\;\; 27=3^3,\;\; 81=3^4,\;\; 243=3^5,\;\ldots
\displaystyle 25=5^2,\;\; 125=5^3,\;\; 625=5^4,\;\ldots

Similarly, one should become familiar with

\displaystyle \frac{1}{4}=\frac{1}{2^2} = 2^{-2},\;\; \frac{1}{8}=\frac{1}{2^3}=2^{-3},\;\; \frac{1}{16}=\frac{1}{2^4}=2^{-4},\;\ldots
\displaystyle \frac{1}{9}=\frac{1}{3^2}=3^{-2},\;\; \frac{1}{27}=\frac{1}{3^3}=3^{-3},\;\ldots
\displaystyle \frac{1}{25}=\frac{1}{5^2}=5^{-2},\;\; \frac{1}{125}=\frac{1}{5^3}=5^{-3},\;\ldots

and so on.

Example 7

  1. Write \displaystyle \ 8^3 \cdot 4^{-2} \cdot 16\ as a power with base 2

    \displaystyle 8^3 \cdot 4^{-2} \cdot 16 = (2^3)^3 \cdot (2^2)^{-2} \cdot 2^4 = 2^{3 \cdot 3} \cdot 2^{2 \cdot (-2)} \cdot 2^4
    \displaystyle \qquad\quad{}= 2^9 \cdot 2^{-4} \cdot 2^4 = 2^{9-4+4} =2^9
  2. Write \displaystyle \ \frac{27^2 \cdot (1/9)^{-2}}{81^2}\ as a power with base 3.

    \displaystyle \frac{27^2 \cdot (1/9)^{-2}}{81^2} = \frac{(3^3)^2 \cdot (1/3^2)^{-2}}{(3^4)^2} = \frac{(3^3)^2 \cdot (3^{-2})^{-2}}{(3^4)^2}
    \displaystyle \qquad\quad{} = \frac{3^{3 \cdot 2} \cdot 3^{(-2) \cdot (-2)}}{3^{4 \cdot 2}} = \frac{3^6\cdot 3^4}{3^8} = \frac{3^{6 + 4}}{3^8}= \frac{3^{10}}{3^8} = 3^{10-8}= 3^2
  3. Write \displaystyle \frac{81 \cdot 32^2 \cdot (2/3)^2}{2^5+2^4} in as simple a form as possible.

    \displaystyle \frac{81 \cdot 32^2 \cdot (2/3)^2}{2^5+2^4} = \frac{3^4 \cdot (2^5)^2 \cdot \displaystyle\frac{2^2}{3^2}}{2^{4+1}+2^4} = \frac{3^4 \cdot 2^{5 \cdot 2} \cdot \displaystyle\frac{2^2}{3^2}}{2^4 \cdot 2^1 +2^4} = \frac{3^4 \cdot 2^{10} \cdot \displaystyle\frac{2^2}{3^2}}{2^4 \cdot(2^1+1)}
    \displaystyle \qquad\quad{} = \frac{ \displaystyle\frac{3^4 \cdot 2^{10} \cdot 2^2}{3^2}}{2^4 \cdot 3} = \frac{ 3^4 \cdot 2^{10} \cdot 2^2 }{3^2 \cdot 2^4 \cdot 3 } = 3^{4-2-1} \cdot 2^{10+2-4} = 3^1 \cdot 2^8= 3\cdot 2^8


Rational exponents

What happens if a number is raised to a rational exponent? Do the definitions and the rules we have used in the above calculations still hold?

For instance we note that

\displaystyle 2^{1/2} \cdot 2^{1/2} = 2^{1/2 + 1/2} = 2^1 = 2

so \displaystyle 2^{1/2} must be the same as \displaystyle \sqrt{2}. This is because \displaystyle \sqrt2 is defined as the number which satisfies \displaystyle \sqrt2\cdot\sqrt2 = 2 .

Generally, we define

\displaystyle a^{1/2} = \sqrt{a}\mbox{.}

We must assume that \displaystyle a\ge 0 since no real number multiplied by itself can give a negative number.

We also see that, for example,

\displaystyle 5^{1/3} \cdot 5^{1/3} \cdot 5^{1/3} = 5^{1/3 + 1/3 +1/3} = 5^1 = 5

which implies that \displaystyle \,5^{1/3} = \sqrt[\scriptstyle3]{5}\mbox{,}\,. This can be generalised to

\displaystyle a^{1/n} = \sqrt[\scriptstyle n]{a}\mbox{.}

By combining this definition with one of our previous laws for exponents, namely \displaystyle ((a^m)^n=a^{m\cdot n}), we have that for all \displaystyle a\ge0, the following holds:

\displaystyle a^{m/n} = (a^m)^{1/n} = \sqrt[\scriptstyle n]{a^m}

or alternatively

\displaystyle a^{m/n} = (a^{1/n})^m = (\sqrt[\scriptstyle n]{a}\,)^m\mbox{.}

Example 8

  1. \displaystyle 27^{1/3} = \sqrt[\scriptstyle 3]{27} = 3\quad as \displaystyle 3 \cdot 3 \cdot 3 =27
  2. \displaystyle 1000^{-1/3} = \frac{1}{1000^{1/3}} = \frac{1}{(10^3)^{1/3}} = \frac{1}{10^{3 \cdot \frac{1}{3}}} = \frac{1}{10^1} = \frac{1}{10}
  3. \displaystyle \frac{1}{\sqrt{8}} = \frac{1}{8^{1/2}} = \frac{1}{(2^3)^{1/2}} = \frac{1}{2^{3/2}} = 2^{-3/2}
  4. \displaystyle \frac{1}{16^{-1/3}} = \frac{1}{(2^4)^{-1/3}} = \frac{1}{2^{-4/3}} = 2^{-(-4/3)}= 2^{4/3}


Comparison of powers

If we do not have access to calculators and wish to compare the size of powers, one can sometimes achieve this by comparing bases or exponents.

If the base of a power is greater than \displaystyle 1 then the power increases as the exponent increases. On the other hand, if the base lies between \displaystyle 0 and \displaystyle 1 then the power decreases as the exponent grows.

Example 9

  1. \displaystyle \quad 3^{5/6} > 3^{3/4}\quad because the base \displaystyle 3 is greater than \displaystyle 1 and the first exponent \displaystyle 5/6 is greater than the second exponent \displaystyle 3/4.
  2. \displaystyle \quad 3^{-3/4} > 3^{-5/6}\quad as the base is greater than \displaystyle 1 and the exponents satisfy \displaystyle -3/4 > - 5/6.
  3. \displaystyle \quad 0\text{.}3^5 < 0\text{.}3^4 \quadas the base \displaystyle 0\text{.}3 is between \displaystyle 0 and \displaystyle 1 and \displaystyle 5 > 4.

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.

Example 10

  1. \displaystyle \quad 5^{3/2} > 4^{3/2}\quad as the base \displaystyle 5 is larger than the base \displaystyle 4 and both powers have the same positive exponent \displaystyle 3/2.
  2. \displaystyle \quad 2^{-5/3} > 3^{-5/3}\quad as the bases satisfy \displaystyle 2<3 and the powers have a negative exponent \displaystyle -5/3.

Sometimes powers need to be rewritten in order to determine the relative sizes. For example to compare \displaystyle 125^2 with \displaystyle 36^3 we can rewrite them as

\displaystyle

125^2 = (5^3)^2 = 5^6\quad \text{and}\quad 36^3 = (6^2)^3 = 6^6

after which we see that \displaystyle 36^3 > 125^2.

Example 11

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

  1. \displaystyle 25^{1/3}   and  \displaystyle 5^{3/4} .

    The base 25 can be rewritten in terms of the second base \displaystyle 5 by putting \displaystyle 25= 5\cdot 5= 5^2. Therefore
    \displaystyle 25^{1/3} = (5^2)^{1/3} = 5^{2 \cdot \frac{1}{3}}= 5^{2/3}\text{,}

    hence we see that

    \displaystyle 5^{3/4} > 25^{1/3}
    since \displaystyle \frac{3}{4} > \frac{2}{3} and the base \displaystyle 5 is larger than \displaystyle 1.
  2. \displaystyle (\sqrt{8}\,)^5   and \displaystyle 128.

    Both \displaystyle 8 and \displaystyle 128 can be written as powers of \displaystyle 2
    \displaystyle \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{.}}

    This gives

    \displaystyle \begin{align*} 
     (\sqrt{8}\,)^5  &= (8^{1/2})^5 = (8)^{5/2} = (2^3)^{5/2}
                  = 2^{3\cdot\frac{5}{2}}= 2^{15/2}\\
 128 &= 2^7 = 2^{14/2}
 \end{align*}

    and thus

    \displaystyle (\sqrt{8}\,)^5 > 128
    because \displaystyle \frac{15}{2} > \frac{14}{2} and the base \displaystyle 2 is greater than \displaystyle 1.
  3. \displaystyle (8^2)^{1/5} and \displaystyle (\sqrt{27}\,)^{4/5}.

    Since \displaystyle 8=2^3 and \displaystyle 27=3^3, the first step is to simplify and write the numbers as powers of \displaystyle 2 and \displaystyle 3 respectively,
    \displaystyle \begin{align*} 
     (8^2)^{1/5} &= (8)^{2/5} = (2^3)^{2/5} = 2^{3\cdot \frac{2}{5}}
              = 2^{6/5}\mbox{,}\\
 (\sqrt{27}\,)^{4/5} &= (27^{1/2})^{4/5}
              = 27^{ \frac{1}{2} \cdot \frac{4}{5}} = 27^{2/5}
              = (3^3)^{2/5} = 3^{3 \cdot \frac{2}{5}}
              = 3^{6/5}\mbox{.}

    \end{align*}

    Now we see that

    \displaystyle (\sqrt{27}\,)^{4/5} > (8^2)^{1/5}

    because \displaystyle 3>2 and exponent \displaystyle \frac{6}{5} is positive.

  4. \displaystyle 3^{1/3}   and  \displaystyle 2^{1/2}

    We rewrite the exponents due to them having a common denominator
    \displaystyle \frac{1}{3} = \frac{2}{6} \quad and \displaystyle \quad \frac{1}{2} = \frac{3}{6}.

    This gives

    \displaystyle \begin{align*} 
     3^{1/3} &= 3^{2/6} = (3^2)^{1/6} = 9^{1/6}\\
 2^{1/2} &= 2^{3/6} = (2^3)^{1/6} = 8^{1/6}

    \end{align*}

    and we see that

    \displaystyle 3^{1/3} > 2^{1/2}
    because \displaystyle 9>8 and the exponent \displaystyle 1/6 is positive.

Exercises


Study advice

Basic and final tests

After you have read the text and worked through the exercises, you should do the basic and final tests to pass this section. You can find the link to the tests in your student lounge.


Keep in mind that...

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


Reviews

For those of you who want to deepen your studies or need more detailed explanations consider the following references

Learn more about powers in the English Wikipedi

What is the greatest prime number? Read more at The Prime Page


Useful web sites

Here you can practise the laws of exponents

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