1.3 Powers

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
m (Robot: Automated text replacement (-Övningar +Exercises))
Current revision (13:57, 26 October 2008) (edit) (undo)
 
(6 intermediate revisions not shown.)
Line 2: Line 2:
{| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"
{| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"
| style="border-bottom:1px solid #797979" width="5px" |  
| style="border-bottom:1px solid #797979" width="5px" |  
-
{{Vald flik|[[1.3 Potenser|Theory]]}}
+
{{Selected tab|[[1.3 Powers|Theory]]}}
-
{{Ej vald flik|[[1.3 Exercises|Exercises]]}}
+
{{Not selected tab|[[1.3 Exercises|Exercises]]}}
| style="border-bottom:1px solid #797979" width="100%"|  
| style="border-bottom:1px solid #797979" width="100%"|  
|}
|}
Line 18: Line 18:
'''Learning outcomes:'''
'''Learning outcomes:'''
-
After this section, you will have learned to:
+
After this section you will have learned to:
* Recognise the concepts of base and exponent.
* Recognise the concepts of base and exponent.
-
*Calculate integer power expressions
+
*Calculate integer power expressions.
*Use the laws of exponents to simplify expressions containing powers.
*Use the laws of exponents to simplify expressions containing powers.
* Know when the laws of exponents are applicable (positive basis).
* Know when the laws of exponents are applicable (positive basis).
Line 29: Line 29:
== Integer exponents ==
== Integer exponents ==
-
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 shorthand 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.
<div class="exempel">
<div class="exempel">
Line 49: Line 49:
= 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>
+
= 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>, but <math> -2^4
= (-2) \cdot (-2) \cdot (-2) \cdot (-2)= 16</math>, but <math> -2^4
Line 79: Line 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>
Line 85: Line 85:
== Laws of exponents ==
== 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
+
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 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\text{,}</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 the division of powers that 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 there is 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>
Line 143: Line 143:
-
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 in both 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>}}
The only way for the rules of exponents to agree is to make the
The only way for the rules of exponents to agree is to make the
-
following but natural definition that for all non zero ''a'' one has that
+
following natural definition. For all non zero ''a'' we have
<div class="regel">
<div class="regel">
-
{{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. For example we have
-
{{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
+
It is therefore necessary that we assume the negative sign of the 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>}}
-
The general definition of negative exponents is to interpret negative exponents
+
We therefore note that the general definition for negative exponents is that for all non zero numbers ''a'', we have
-
of all non zero numbers ''a'' as follows
+
of 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>
Line 190: Line 190:
</div>
</div>
-
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>
if <math>n</math> is odd and equal to <math>+1</math> if <math>n</math> is even .
if <math>n</math> is odd and equal to <math>+1</math> if <math>n</math> is even .
Line 215: Line 215:
==Changing the base ==
==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
+
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:
-
{{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
+
Similarly, one should become familiar with
-
{{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.
Line 260: Line 260:
== Rational exponents ==
== 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?
+
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, since
+
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>}}
-
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>. This is because <math>\sqrt2</math> is defined as the number which satisfies <math>\sqrt2\cdot\sqrt2 = 2</math>&nbsp;.
Generally, we define
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>
-
We must assume that <math>a\ge 0</math>, since no real number multiplied by itself can give a negative number.
+
We must assume that <math>a\ge 0</math> since no real number multiplied by itself can give a negative number.
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 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>
-
By combining this definition with one of the previous laws of exponents <math>((a^m)^n=a^{m\cdot n})</math> gives that for all <math>a\ge0</math> it holds that
+
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>}}
+
{{Displayed math||<math>a^{m/n} = (a^m)^{1/n} = \sqrt[\scriptstyle n]{a^m}</math>}}
-
or
+
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>
Line 316: Line 316:
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 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 <math>1</math> 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.
+
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">
Line 322: Line 322:
<ol type="a">
<ol type="a">
-
<li><math>\quad 3^{5/6} > 3^{3/4}\quad</math> as the base <math>3</math> is greater than <math>1</math> and the first exponent <math>5/6</math> is greater than the second exponent <math>3/4</math>.</li>
+
<li><math>\quad 3^{5/6} > 3^{3/4}\quad</math> because the base <math>3</math> is greater than <math>1</math> and the first exponent <math>5/6</math> is greater than the second exponent <math>3/4</math>.</li>
<li><math>\quad 3^{-3/4} > 3^{-5/6}\quad</math> as the base is greater than <math>1</math> and the exponents satisfy <math> -3/4 > - 5/6</math>.</li>
<li><math>\quad 3^{-3/4} > 3^{-5/6}\quad</math> as the base is greater than <math>1</math> and the exponents satisfy <math> -3/4 > - 5/6</math>.</li>
-
<li><math> \quad 0{,}3^5 < 0{,}3^4 \quad</math>as the base <math> 0{,}3</math> is between <math>0</math> and <math>1</math> and <math>5 > 4</math>.
+
<li><math> \quad 0\text{.}3^5 < 0\text{.}3^4 \quad</math>as the base <math> 0\text{.}3</math> is between <math>0</math> and <math>1</math> and <math>5 > 4</math>.
</ol>
</ol>
</div>
</div>
-
If a power has a positive exponent, it will get larger the larger the base becomes. The opposite applies if the exponent is negative: that is, the power decreases as the base gets larger.
+
If a power has a positive exponent it 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">
Line 339: Line 339:
</div>
</div>
-
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 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{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>}}
-
after which one can see that <math>36^3 > 125^2</math>.
+
after which we see that <math>36^3 > 125^2</math>.
<div class="exempel">
<div class="exempel">
''' Example 11'''
''' Example 11'''
-
Determine which of the following pairs of numbers is the greater
+
Determine which of the following pairs of numbers is the greater:
<ol type="a">
<ol type="a">
Line 357: Line 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}\text{,}</math>}}
-
and then we see that
+
hence 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>
Line 370: Line 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 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 382: Line 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>
Line 389: Line 389:
<br>
<br>
<br>
<br>
-
Since <math>8=2^3</math> and <math>27=3^3</math> a first step can be to simplify and write the numbers as powers of <math>2</math> and <math>3</math> respectively,
+
Since <math>8=2^3</math> and <math>27=3^3</math>, the first step is to simplify and write the numbers as powers of <math>2</math> and <math>3</math> respectively,
-
{{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 402: Line 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.
Line 409: Line 409:
<br>
<br>
<br>
<br>
-
We rewrite the exponents so they have a common denominator
+
We rewrite the exponents due to them having a common denominator
-
{{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
+
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}
Line 422: Line 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>
Line 439: Line 439:
-
'''Keep in mind that:'''
+
'''Keep in mind that...'''
-
The number raised to the power 0, is always 1, if the number (the base) is not 0.
+
The number raised to the power 0 is always 1 as long as the number (the base) is not 0.

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