1.3 Potenzen

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Zeile 31: Zeile 31:
We use the multiplication symbol as a short-hand for repeated addition of the same number, for example,
We use the multiplication symbol as a short-hand for repeated addition of the same number, for example,
-
{{Fristående formel||<math>4 + 4 + 4 + 4 + 4 = 4 \cdot 5\mbox{.}</math>}}
+
{{Displayed math||<math>4 + 4 + 4 + 4 + 4 = 4 \cdot 5\mbox{.}</math>}}
In a similar way we use exponentials as a short-hand for repeated multiplication
In a similar way we use exponentials as a short-hand for repeated multiplication
of the same number:
of the same number:
-
{{Fristående formel||<math> 4 \cdot 4 \cdot 4 \cdot 4 \cdot 4 = 4^5\mbox{.}</math>}}
+
{{Displayed math||<math> 4 \cdot 4 \cdot 4 \cdot 4 \cdot 4 = 4^5\mbox{.}</math>}}
The 4 is called the base of the power, and the 5 is its exponent.
The 4 is called the base of the power, and the 5 is its exponent.
Zeile 79: Zeile 79:
<div class="regel">
<div class="regel">
-
{{Fristående formel||<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>}}
+
{{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>
Zeile 87: Zeile 87:
There are a few more rules coming from the definition of power which are useful when doing calculations.You can see for example that
There are a few more rules coming from the definition of power which are useful when doing calculations.You can see for example that
-
{{Fristående formel||<math>2^3 \cdot 2^5 = \underbrace{\,2\cdot 2\cdot 2\vphantom{{}_{\scriptscriptstyle 1}}\,}_{ 3\ {\rm factors }} \cdot \underbrace{\,2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\vphantom{{}_{\scriptscriptstyle 1}}\,}_{ 5\ {\rm factors }} = \underbrace{\,2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\vphantom{{}_{\scriptscriptstyle 1}}\,}_{ (3 + 5)\ {\rm factors}} = 2^{3+5} = 2^8</math>}}
+
{{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</math>}}
which generally can be expressed as
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>}}
+
{{Displayed math||<math>a^m \cdot a^n = a^{m+n}\mbox{.}</math>}}
</div>
</div>
There is also a useful simplification rule for division of powers which have the same base.
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>}}
+
{{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>}}
The general rule is
The general rule is
<div class="regel">
<div class="regel">
-
{{Fristående formel||<math>\displaystyle\frac{a^m}{a^n}= a^{m-n}\mbox{.}</math>}}
+
{{Displayed math||<math>\displaystyle\frac{a^m}{a^n}= a^{m-n}\mbox{.}</math>}}
</div>
</div>
For the case when the base itself is a power one has another useful rule. We see that
For the case when the base itself is a power one has another useful rule. We see that
-
{{Fristående formel||<math> (5^2)^3 = 5^2 \cdot 5^2 \cdot 5^2 = \underbrace{\,5\cdot 5\vphantom{{}_{\scriptscriptstyle 1}}\,}_{ 2\ {\rm 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>}}
+
{{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>}}
and
and
-
{{Fristående formel||<math> (5^3)^2 = 5^3\cdot5^3= \underbrace{\,5\cdot 5 \cdot 5\vphantom{{}_{\scriptscriptstyle 1}}\,}_{3\ {\rm factors}} \cdot \underbrace{\,5\cdot 5 \cdot 5\vphantom{{}_{\scriptscriptstyle 1}}\,}_{3\ {\rm factors}} = \underbrace{\,5\cdot 5 \cdot 5\,\cdot\,5\cdot 5 \cdot 5\vphantom{{}_{\scriptscriptstyle 1}}\,}_{2\ {\rm times}\ 3\ {\rm factors}}=5^{3\cdot2}=5^6\mbox{.}</math>}}
+
{{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>}}
Generally, this can be written
Generally, this can be written
<div class="regel">
<div class="regel">
-
{{Fristående formel||<math>(a^m)^n = a^{m \cdot n}\mbox{.} </math>}}
+
{{Displayed math||<math>(a^m)^n = a^{m \cdot n}\mbox{.} </math>}}
</div>
</div>
Zeile 144: Zeile 144:
If a fraction has the same expression for the exponent both in the numerator and the denominator we can simplify in two ways:
If a fraction has the same expression for the exponent 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{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>}}
+
{{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>}}
Zeile 152: Zeile 152:
<div class="regel">
<div class="regel">
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{{Fristående formel||<math> a^0 = 1\mbox{.} </math>}}
+
{{Displayed math||<math> a^0 = 1\mbox{.} </math>}}
</div>
</div>
We can also run into examples where the exponent in the denominator is greater than that in the numerator. We can have, for example,
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{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>}}
+
{{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>}}
We see that it is necessary to assume that the negative exponent implies that
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>}}
+
{{Displayed math||<math>3^{-2} = \frac{1}{3^2}\mbox{.}</math>}}
Zeile 165: Zeile 165:
of all non zero numbers ''a'' as follows
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>}}
+
{{Displayed math||<math>a^{-n} = \frac{1}{a^n}\mbox{.}</math>}}
</div>
</div>
Zeile 192: Zeile 192:
If the base of a power is <math>-1</math> then the expression will simplify to either <math>-1</math> or <math>+1</math> depending on the value of the exponent
If the base of a power is <math>-1</math> then the expression will simplify to either <math>-1</math> or <math>+1</math> depending on the value of the exponent
-
{{Fristående formel||<math>\eqalign{(-1)^1 &= -1\cr (-1)^2 &= (-1)\cdot(-1) = +1\cr (-1)^3 &= (-1)\cdot(-1)^2 = (-1)\cdot 1 = -1\cr (-1)^4 &= (-1)\cdot(-1)^3 = (-1)\cdot (-1) = +1\cr \quad\hbox{etc.}}</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>}}
The rule is that <math>(-1)^n </math> is equal to<math>-1</math>
The rule is that <math>(-1)^n </math> is equal to<math>-1</math>
Zeile 217: Zeile 217:
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
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>}}
+
{{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>}}
+
{{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>}}
+
{{Displayed math||<math>25=5^2,\;\; 125=5^3,\;\; 625=5^4,\;\ldots</math>}}
But even
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>}}
+
{{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>}}
and so on.
and so on.
Zeile 263: Zeile 263:
For instance, since
For instance, since
-
{{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>}}
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;.
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;.
Zeile 269: Zeile 269:
<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>
Zeile 275: Zeile 275:
We also see that, for example,
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>}}
which means that <math>\,5^{1/3} = \sqrt[\scriptstyle3]{5}\mbox{,}\,</math> which can be generalised to
which means that <math>\,5^{1/3} = \sqrt[\scriptstyle3]{5}\mbox{,}\,</math> which 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>
Zeile 286: Zeile 286:
<div class="regel">
<div class="regel">
-
{{Fristående formel||<math>a^{m/n} = (a^m)^{1/n} = \sqrt[\scriptstyle n]{a^m}</math>}}
+
{{Displayed math||<math>a^{m/n} = (a^m)^{1/n} = \sqrt[\scriptstyle n]{a^m}</math>}}
or
or
-
{{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>
Zeile 340: Zeile 340:
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
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>
+
{{Displayed math||<math>
125^2 = (5^3)^2 = 5^6\quad \text{and}\quad 36^3 = (6^2)^3 = 6^6
125^2 = (5^3)^2 = 5^6\quad \text{and}\quad 36^3 = (6^2)^3 = 6^6
</math>}}
</math>}}
Zeile 357: Zeile 357:
The base 25 can be rewritten in terms of the second base <math>5</math> by putting <math>25= 5\cdot 5= 5^2</math>. Therefore
The base 25 can be rewritten in terms of the second base <math>5</math> by putting <math>25= 5\cdot 5= 5^2</math>. Therefore
-
{{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}</math>}}
and then we see that
and then we see that
-
{{Fristående formel||<math>5^{3/4} > 25^{1/3} </math>}}
+
{{Displayed math||<math>5^{3/4} > 25^{1/3} </math>}}
since <math>\frac{3}{4} > \frac{2}{3}</math> and the base <math>5</math> is larger than <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>
Zeile 370: Zeile 370:
Both <math>8</math> and <math>128</math> can be written as powers of <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>}}
This means that
This means that
-
{{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}\\
Zeile 382: Zeile 382:
and thus
and thus
-
{{Fristående formel||<math>(\sqrt{8}\,)^5 > 128 </math>}}
+
{{Displayed math||<math>(\sqrt{8}\,)^5 > 128 </math>}}
because <math>\frac{15}{2} > \frac{14}{2}</math> and the base <math>2</math> is greater than <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>
Zeile 391: Zeile 391:
Since <math>8=2^3</math> and <math>27=3^3</math> a first step can be to simplify and write the numbers as powers of <math>2</math> and <math>3</math> respectively,
Since <math>8=2^3</math> and <math>27=3^3</math> 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*}
+
{{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{,}\\
Zeile 402: Zeile 402:
Now we see that
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>}}
because <math> 3>2</math> and exponent <math>\frac{6}{5}</math> is positive.
because <math> 3>2</math> and exponent <math>\frac{6}{5}</math> is positive.
Zeile 411: Zeile 411:
We rewrite the exponents so they have a common denominator
We rewrite the exponents so they have a common denominator
-
{{Fristående formel||<math>\frac{1}{3} = \frac{2}{6} \quad</math> and <math>\quad \frac{1}{2} = \frac{3}{6}</math>.}}
+
{{Displayed math||<math>\frac{1}{3} = \frac{2}{6} \quad</math> and <math>\quad \frac{1}{2} = \frac{3}{6}</math>.}}
Then we have that
Then we have that
-
{{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}
Zeile 422: Zeile 422:
and we see that
and we see that
-
{{Fristående formel||<math> 3^{1/3} > 2^{1/2} </math>}}
+
{{Displayed math||<math> 3^{1/3} > 2^{1/2} </math>}}
because <math> 9>8</math> and the exponent <math>1/6</math> is positive.</li>
because <math> 9>8</math> and the exponent <math>1/6</math> is positive.</li>

Version vom 13:54, 10. Sep. 2008

 

Vorlage:Selected tab Vorlage:Not selected tab

 

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 short-hand for repeated addition of the same number, for example,

Vorlage:Displayed math

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

Vorlage:Displayed math

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, 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:

Vorlage:Displayed math


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

Vorlage:Displayed math

which generally can be expressed as

Vorlage:Displayed math

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

Vorlage:Displayed math

The general rule is

Vorlage:Displayed math

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

Vorlage:Displayed math

and

Vorlage:Displayed math


Generally, this can be written

Vorlage:Displayed math

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 both in the numerator and the denominator we can simplify in two ways: Vorlage:Displayed math


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


Vorlage:Displayed math

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

We see that it is necessary to assume that the negative exponent implies that Vorlage:Displayed math


The general definition of negative exponents is to interpret negative exponents of all non zero numbers a as follows

Vorlage:Displayed math


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

Vorlage:Displayed math

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 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

Vorlage:Displayed math

Vorlage:Displayed math

Vorlage:Displayed math

But even

Vorlage:Displayed math

Vorlage:Displayed math

Vorlage:Displayed math

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 above to do calculations still hold?

For instance, since Vorlage:Displayed math 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

Vorlage:Displayed math

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, Vorlage:Displayed math

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

Vorlage:Displayed math

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

Vorlage:Displayed math

or

Vorlage:Displayed math

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 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 get larger the larger the base becomes. The opposite applies if the exponent is negative: that is, the power decreases as the base gets larger.

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 Vorlage:Displayed math

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}   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 Vorlage:Displayed math and then we see that Vorlage:Displayed math 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 Vorlage:Displayed math This means that Vorlage:Displayed math and thus Vorlage:Displayed math 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 a first step can be to simplify and write the numbers as powers of \displaystyle 2 and \displaystyle 3 respectively, Vorlage:Displayed math Now we see that Vorlage:Displayed math 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 so they have a common denominator Vorlage:Displayed math Then we have that Vorlage:Displayed math and we see that Vorlage:Displayed math 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


Useful web sites

Here you can practise the laws of exponents

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/1.3_Potenzen“