1.3 Powers

From Förberedande kurs i matematik 1

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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
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{{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 containg 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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*Determining which of two powers is the largest based on a comparison of the
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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 short-hand 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>}}
{{Fristående formel||<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
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of the same number:
{{Fristående formel||<math> 4 \cdot 4 \cdot 4 \cdot 4 \cdot 4 = 4^5\mbox{.}</math>}}
{{Fristående formel||<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 '''
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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:
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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 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 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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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>}}
{{Fristående formel||<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 division of powers which have the same base.
{{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>}}
{{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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Den allmänna regeln blir
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The general rule is
<div class="regel">
<div class="regel">
{{Fristående formel||<math>\displaystyle\frac{a^m}{a^n}= a^{m-n}\mbox{.}</math>}}
{{Fristående formel||<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 that is the power of a power one has 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>}}
{{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>}}
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och
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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 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>}}
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Allmänt kan detta skrivas
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Generally, this can be written
<div class="regel">
<div class="regel">
{{Fristående formel||<math>(a^m)^n = a^{m \cdot n}\mbox{.} </math>}}
{{Fristående formel||<math>(a^m)^n = a^{m \cdot n}\mbox{.} </math>}}
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<div class="exempel">
<div class="exempel">
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'''Exempel 3'''
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''' Example 3'''
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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 power (expression) both in the numerator and the denominator we can simplify in two ways:
{{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>}}
{{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>}}
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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
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The only way for the rules of powers to agree is to make the
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following but natural definition that for all non zero ''a'' one has that
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<div class="regel">
<div class="regel">
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Vi kan också råka ut för att exponenten i nämnaren är större än den i täljaren. Vi får t.ex.
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We can also run into examples where the exponent in the denominator is greater than that in the numerator. We can have, for example,
{{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>}}
{{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>}}
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Vi ser här att enligt våra räkneregler måste den negativa exponenten betyda att
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We see that it is necessary to assume that the negative exponent implies that
{{Fristående formel||<math>3^{-2} = \frac{1}{3^2}\mbox{.}</math>}}
{{Fristående formel||<math>3^{-2} = \frac{1}{3^2}\mbox{.}</math>}}
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Den allmänna definitionen av negativa exponenter är att, för alla tal ''a'' som inte är 0 gäller att
 
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The general definition of negative exponents is to interpret negative exponents
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of all non zero numbers ''a'' as follows
<div class="regel">
<div class="regel">
{{Fristående formel||<math>a^{-n} = \frac{1}{a^n}\mbox{.}</math>}}
{{Fristående formel||<math>a^{-n} = \frac{1}{a^n}\mbox{.}</math>}}
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<div class="exempel">
<div class="exempel">
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'''Exempel 5'''
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''' Example 5'''
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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> s 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{osv.}}</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>
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if <math>n</math>is odd and equal to <math>+1</math> if <math>n</math>is even .
<div class="exempel">
<div class="exempel">
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'''Exempel 6'''
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''' Example 6'''
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<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>
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<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 try, if possible, to combine powers by choosing the same base. This often involves selecting 2, 3 or 5 as a base and, therefore, it is a good idea to learn to recognize the powers of these numbers, such as
{{Fristående formel||<math>4=2^2,\;\; 8=2^3,\;\; 16=2^4,\;\; 32=2^5,\;\; 64=2^6,\;\; 128=2^7,\;\ldots</math>}}
{{Fristående formel||<math>4=2^2,\;\; 8=2^3,\;\; 16=2^4,\;\; 32=2^5,\;\; 64=2^6,\;\; 128=2^7,\;\ldots</math>}}
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{{Fristående formel||<math>25=5^2,\;\; 125=5^3,\;\; 625=5^4,\;\ldots</math>}}
{{Fristående formel||<math>25=5^2,\;\; 125=5^3,\;\; 625=5^4,\;\ldots</math>}}
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Men även
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But even
{{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>}}
{{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>}}
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{{Fristående formel||<math>\frac{1}{25}=\frac{1}{5^2}=5^{-2},\;\; \frac{1}{125}=\frac{1}{5^3}=5^{-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>}}
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osv.
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and so on.
<div class="exempel">
<div class="exempel">
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'''Exempel 7'''
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''' Example 7'''
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<li>Skriv <math>\ 8^3 \cdot 4^{-2} \cdot 16\ </math> som en potens med basen 2.
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<li> Write <math>\ 8^3 \cdot 4^{-2} \cdot 16\ </math> s 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>
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<li>Skriv <math>\ \frac{27^2 \cdot (1/9)^{-2}}{81^2}\ </math> som en potens av basen 3.
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<li> Write <math>\ \frac{27^2 \cdot (1/9)^{-2}}{81^2}\ </math> as a power with base 3.
<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>
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<li>Skriv <math>\frac{81 \cdot 32^2 \cdot (2/3)^2}{2^5+2^4}</math> så enkelt som möjligt.
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<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 ==
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== Rational exponents ==
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Vad händer om ett tal höjs upp till en rationell exponent? Gäller fortfarande de definitioner och räkneregler vi har använt oss av ovan?
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What happens if a number is raised to a rational exponent? Do the the definitions and the rules we have used above to do calculations still hold?
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Eftersom exempelvis
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For instance, since
{{Fristående formel||<math>2^{1/2} \cdot 2^{1/2} = 2^{1/2 + 1/2} = 2^1 = 2</math>}}
{{Fristående formel||<math>2^{1/2} \cdot 2^{1/2} = 2^{1/2 + 1/2} = 2^1 = 2</math>}}
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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;.
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so <math> 2^{1/2} </math> must be the same as <math>\sqrt{2}</math> because <math>\sqrt2</math> is defined as the number which satisfies <math>\sqrt2\cdot\sqrt2 = 2</math>&nbsp;.
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Allmänt kan vi göra definitionen
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Generally, we define
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</div>
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Vi måste då förutsätta att <math>a\ge 0</math>, eftersom inget reellt tal multiplicerat med sig själv kan ge ett negativt tal.
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We must assume that t <math>a\ge 0</math>, ince no real number multiplied by itself can give a negative number.
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Man ser också att exempelvis
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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>}}
{{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>}}
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som innebär att <math>\,5^{1/3} = \sqrt[\scriptstyle3]{5}\mbox{,}\,</math> vilket kan generaliseras till att
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which means that math>\,5^{1/3} = \sqrt[\scriptstyle3]{5}\mbox{,}\,</math> which can be generalised to
<div class="regel">
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</div>
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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
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By combining this definition with one of the previous laws of exponents <math>((a^m)^n=a^{m\cdot n})</math> means that, for all<math>a\ge0</math> it holds that
<div class="regel">
<div class="regel">
{{Fristående formel||<math>a^{m/n} = (a^m)^{1/n} = \sqrt[\scriptstyle n]{a^m}</math>}}
{{Fristående formel||<math>a^{m/n} = (a^m)^{1/n} = \sqrt[\scriptstyle n]{a^m}</math>}}
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eller
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or
{{Fristående formel||<math>a^{m/n} = (a^{1/n})^m = (\sqrt[\scriptstyle n]{a}\,)^m\mbox{.} </math>}}
{{Fristående formel||<math>a^{m/n} = (a^{1/n})^m = (\sqrt[\scriptstyle n]{a}\,)^m\mbox{.} </math>}}
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<div class="exempel">
<div class="exempel">
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'''Exempel 8'''
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''' Example 8'''
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== Jämförelse av potenser ==
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==Comparison of powers ==
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Om man utan tillgång till miniräknare vill jämföra storleken av potenser, kan man i vissa fall avgöra detta genom att jämföra basen eller exponenten.
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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.
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Om basen i en potens är större 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.
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If the base of a power greater than <math>1</math> s then the power is larger the larger the exponent. 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">
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'''Exempel 9'''
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''' Example 9'''
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<ol type="a">
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<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>
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<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>
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<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>
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<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>
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<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>.
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<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>.
</ol>
</ol>
</div>
</div>
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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.
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If a power has a positive exponent, it will larger the larger the base is . The opposite applies if exponent is negative: that is the power decreases as the base gets larger.
<div class="exempel">
<div class="exempel">
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'''Exempel 10'''
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''' Example 10'''
<ol type="a">
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<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 must 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
{{Fristående formel||<math>
{{Fristående formel||<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{och}\quad 36^3 = (6^2)^3 = 6^6
</math>}}
</math>}}
-
varefter man kan konstatera att <math>36^3 > 125^2</math>.
+
after which one can 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
+
Determine which of the following pairs of numbers is the greater
<ol type="a">
<ol type="a">
Line 352: Line 356:
<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 written about 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>}}
{{Fristående formel||<math>25^{1/3} = (5^2)^{1/3} = 5^{2 \cdot \frac{1}{3}}= 5^{2/3}</math>}}
-
och då ser vi att
+
  and then we see that
{{Fristående formel||<math>5^{3/4} > 25^{1/3} </math>}}
{{Fristående formel||<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; och <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>}}
{{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>}}
-
Detta betyder att
+
This means that
{{Fristående formel||<math>\begin{align*}
{{Fristående formel||<math>\begin{align*}
Line 377: Line 381:
\end{align*}</math>}}
\end{align*}</math>}}
-
och därför är
+
and thus
{{Fristående formel||<math>(\sqrt{8}\,)^5 > 128 </math>}}
{{Fristående formel||<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> och <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> a first step can be to simplify and write the numbers as powers of <math>2</math> and <math>3</math> respectively,
{{Fristående formel||<math>\begin{align*}
{{Fristående formel||<math>\begin{align*}
Line 397: Line 401:
\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>}}
{{Fristående formel||<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> nd 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; och &nbsp;<math> 2^{1/2}</math>
<br>
<br>
<br>
<br>
-
Vi skriver exponenterna med gemensam nämnare
+
We rewrite the exponents so they have a common denominator
-
{{Fristående formel||<math>\frac{1}{3} = \frac{2}{6} \quad</math> och <math>\quad \frac{1}{2} = \frac{3}{6}</math>.}}
+
{{Fristående formel||<math>\frac{1}{3} = \frac{2}{6} \quad</math> and <math>\quad \frac{1}{2} = \frac{3}{6}</math>.}}
-
Då har vi att
+
Then we have that
{{Fristående formel||<math>\begin{align*}
{{Fristående formel||<math>\begin{align*}
Line 417: Line 421:
\end{align*}</math>}}
\end{align*}</math>}}
-
och vi ser att
+
  and we see that
{{Fristående formel||<math> 3^{1/3} > 2^{1/2} </math>}}
{{Fristående formel||<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 Övningar|Exercises]]
<div class="inforuta" style="width:580px;">
<div class="inforuta" style="width:580px;">
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'''Råd för inläsning'''
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'''Study advice''
-
'''Grund- och slutprov'''
+
'''Basic and final tests'''
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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:'''
+
'''Keep in mind that:'''
-
Ett tal upphöjt till 0 är 1, om talet (basen) är skild från 0.
+
The number raised to the power 0, is always 1, if the number (the base) is not 0.
-
'''Lästips'''
+
'''Reviews'''
-
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
-
[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'''
'''Länktips'''
-
[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>

Revision as of 12:51, 9 July 2008

       Teori          Övningar      

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 containg powers.
  • Know when the laws of exponents are applicable (positive basis).
  • Determining which of two powers is the largest based on a comparison of the

base / exponent.

Integer exponents

We use the multiplication symbol as a short-hand 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{,}1 \cdot 0{,}1 \cdot 0{,}1 = 0{,}001
  4. \displaystyle (-2)^4 = (-2) \cdot (-2) \cdot (-2) \cdot (-2)= 16, men \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, men \displaystyle (2\cdot3)^2 = 6^2 = 36

Exempel 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{och}\quad (ab)^m = a^m b^m\,\mbox{.}


Laws of exponents

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

\displaystyle 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

which generally can be expressed as

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

There is also a useful simplification rule for division of powers which 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 that is the power of a power one has 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 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{.}

and

\displaystyle (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{.}


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

Exempel 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 power (expression) both in the numerator and the denominator we can simplify in two ways:

\displaystyle \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{.}


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


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

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

\displaystyle \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{.}

We see that it is necessary to assume that the negative exponent implies that

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


The general definition of negative exponents is to interpret negative exponents of all non zero numbers a 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 s 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{osv.}}

The rule is that \displaystyle (-1)^n is equal to\displaystyle -1 if \displaystyle nis odd and equal to \displaystyle +1 if \displaystyle nis 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 try, if possible, to combine powers by choosing the same base. This often involves selecting 2, 3 or 5 as a base and, therefore, it is a good idea to learn to recognize 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

But even

\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\ s 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 the definitions and the rules we have used above to do calculations still hold?

For instance, since

\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} 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 t \displaystyle a\ge 0, ince 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 means that math>\,5^{1/3} = \sqrt[\scriptstyle3]{5}\mbox{,}\,</math> which can be generalised to

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

By combining this definition with one of the previous laws of exponents \displaystyle ((a^m)^n=a^{m\cdot n}) means that, for all\displaystyle a\ge0 it holds that

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

or

\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 eftersom \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 greater than \displaystyle 1 s then the power is larger the larger the exponent. 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 as 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{,}3^5 < 0{,}3^4 \quadas the base \displaystyle 0{,}3 is between \displaystyle 0 and \displaystyle 1 and \displaystyle 5 > 4.

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

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

\displaystyle

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

after which one can 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}   och  \displaystyle 5^{3/4} .

    The base 25 can be written about in terms of the second base \displaystyle 5by 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}

      and then 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   och \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 means that

    \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} och \displaystyle (\sqrt{27}\,)^{4/5}.

    Since \displaystyle 8=2^3 and \displaystyle 27=3^3 a first step can be 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 nd exponent\displaystyle \frac{6}{5} is positive.

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

    We rewrite the exponents so they have a common denominator
    \displaystyle \frac{1}{3} = \frac{2}{6} \quad and \displaystyle \quad \frac{1}{2} = \frac{3}{6}.

    Then we have that

    \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, if 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


Länktips

Here you can practise the laws of exponents

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