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1.3 Powers

From Förberedande kurs i matematik 1

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= 0{,}1 \cdot 0{,}1 \cdot 0{,}1 = 0{,}001</math></li>
= 0{,}1 \cdot 0{,}1 \cdot 0{,}1 = 0{,}001</math></li>
<li><math>(-2)^4
<li><math>(-2)^4
-
= (-2) \cdot (-2) \cdot (-2) \cdot (-2)= 16</math>, men <math> -2^4
+
= (-2) \cdot (-2) \cdot (-2) \cdot (-2)= 16</math>, but <math> -2^4
= -(2^4) = - (2 \cdot 2 \cdot 2 \cdot 2) = -16</math></li>
= -(2^4) = - (2 \cdot 2 \cdot 2 \cdot 2) = -16</math></li>
-
<li><math> 2\cdot 3^2 = 2 \cdot 3 \cdot 3 = 18</math>, men <math>
+
<li><math> 2\cdot 3^2 = 2 \cdot 3 \cdot 3 = 18</math>, but <math>
(2\cdot3)^2 = 6^2 = 36</math></li>
(2\cdot3)^2 = 6^2 = 36</math></li>
</ol>
</ol>
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<div class="regel">
<div class="regel">
-
{{Fristående formel||<math>\left(\displaystyle\frac{a}{b}\right)^m = \displaystyle\frac{a^m}{b^m} \quad \mbox{och}\quad (ab)^m = a^m b^m\,\mbox{.}</math>}}
+
{{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>}}
</div>
</div>
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{{Fristående formel||<math>\eqalign{(-1)^1 &= -1\cr (-1)^2 &= (-1)\cdot(-1) = +1\cr (-1)^3 &= (-1)\cdot(-1)^2 = (-1)\cdot 1 = -1\cr (-1)^4 &= (-1)\cdot(-1)^3 = (-1)\cdot (-1) = +1\cr \quad\hbox{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>}}
-
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 .
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</div>
</div>
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We must assume that t <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,
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</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 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
<div class="regel">
<div class="regel">
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<ol type="a">
<ol type="a">
<li><math>27^{1/3} = \sqrt[\scriptstyle 3]{27}
<li><math>27^{1/3} = \sqrt[\scriptstyle 3]{27}
-
= 3\quad</math> eftersom <math>3 \cdot 3 \cdot 3 =27</math></li>
+
= 3\quad</math> as <math>3 \cdot 3 \cdot 3 =27</math></li>
<li><math>1000^{-1/3} = \frac{1}{1000^{1/3}}
<li><math>1000^{-1/3} = \frac{1}{1000^{1/3}}
= \frac{1}{(10^3)^{1/3}}
= \frac{1}{(10^3)^{1/3}}
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<ol type="a">
<ol type="a">
-
<li><math> 25^{1/3} </math>&nbsp; och &nbsp;<math> 5^{3/4} </math>.
+
<li><math> 25^{1/3} </math>&nbsp; and &nbsp;<math> 5^{3/4} </math>.
<br>
<br>
<br>
<br>
-
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
+
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>}}
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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>
-
<li><math>(\sqrt{8}\,)^5 </math>&nbsp; och <math>128</math>.
+
<li><math>(\sqrt{8}\,)^5 </math>&nbsp; and <math>128</math>.
<br>
<br>
<br>
<br>
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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>
-
<li><math> (8^2)^{1/5} </math> och <math> (\sqrt{27}\,)^{4/5}</math>.
+
<li><math> (8^2)^{1/5} </math> and <math> (\sqrt{27}\,)^{4/5}</math>.
<br>
<br>
<br>
<br>
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{{Fristående formel||<math>(\sqrt{27}\,)^{4/5} > (8^2)^{1/5} </math>}}
{{Fristående formel||<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.
-
<li><math> 3^{1/3} </math>&nbsp; och &nbsp;<math> 2^{1/2}</math>
+
<li><math> 3^{1/3} </math>&nbsp; and &nbsp;<math> 2^{1/2}</math>
<br>
<br>
<br>
<br>
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{{Fristående formel||<math> 3^{1/3} > 2^{1/2} </math>}}
{{Fristående formel||<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>
</ol>
</ol>
</div>
</div>

Revision as of 17:15, 6 August 2008

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

4+4+4+4+4=45.

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

44444=45.

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

Example 1

  1. 53=555=125
  2. 105=1010101010=100000
  3. 013=010101=0001
  4. (−2)4=(−2)(−2)(−2)(−2)=16, but −24=−(24)=−(2222)=−16
  5. 232=233=18, but (23)2=62=36

Exempel 2

  1. 323=323232=3323=827 
  2. (23)4=(23)(23)(23)(23)
    =22223333=2434=1296

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

bam=bmamand(ab)m=ambm. 


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

2325=2223 st2222225 st=22222222(3+5) st=23+5=28

which generally can be expressed as

aman=am+n.

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

2327=2222222222=27−3=24.

The general rule is

anam=am−n.

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

(52)3=525252=552 st552 st552 st=5555553 gånger 2 st=523=56.

and

(53)2=5353=5553 st5553 st=5555552 gånger 3 st=532=56.


Generally, this can be written

(am)n=amn.

Example 3

  1. 29214=29+14=223
  2. 553=5153=51+3=54
  3. 323334=32+3+4=39
  4. 1051000=105103=105+3=108

Exempel 4

  1. 3983100=3100−98=32
  2. 7710=71710=710−1=79


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

5353=53−3=50as well as5353=555555=125125=1.


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


a0=1.

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

3634=34−6=3−2and3634=3333333333=133=132.

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

3−2=132.


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

a−n=1an.


Example 5

  1. 7129371293=71293−1293=70=1
  2. 373−934=37+(−9)+4=32
  3. 0001=11000=1103=10−3
  4. 0008=81000=1125=153=5−3
  5. 32−1=1321=123=23
  6. 132−3=(3−2)−3=3(−2)(−3)=36 
  7. 0015=(10−2)5=10−25=10−10

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

(−1)1(−1)2(−1)3(−1)4osv.=−1=(−1)(−1)=+1=(−1)(−1)2=(−1)1=−1=(−1)(−1)3=(−1)(−1)=+1

The rule is that (−1)n is equal to−1 if n is odd and equal to +1 if n is even .


Example 6

  1. (−1)56=1 as 56 is an even number
  2. 1(−1)11=1−1=−1 because 11 is an odd number
  3. 2130(−2)127=2130(−12)127=2130(−1)1272127=2130−12127 =−2127−130=−2−3=−123=−81


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

4=228=2316=2432=2564=26128=27...
9=3227=3381=34243=35...
25=52125=53625=54...

But even

41=122=2−281=123=2−3116=124=2−4...
91=132=3−2127=133=3−3...
125=152=5−21125=153=5−3...

and so on.

Example 7

  1. Write  834−216  as a power with base 2

    834−216=(23)3(22)−224=23322(−2)24
    =292−424=29−4+4=29
  2. Write  812272(19)−2  as a power with base 3.

    812272(19)−2=(34)2(33)2(132)−2=(34)2(33)2(3−2)−2
    =3423323(−2)(−2)=383634=3836+4=38310=310−8=32
  3. Write 25+2481322(23)2 in as simple a form as possible.

    25+2481322(23)2=24+1+2434(25)23222=2421+24342523222=24(21+1)342103222
    =243323421022=322433421022=34−2−1210+2−4=3128=328


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

212212=212+12=21=2 

so 212  must be the same as 2  because 2  is defined as the number which satisfies 22=2  .

Generally, we define

a12=a. 

We must assume that a0, since no real number multiplied by itself can give a negative number.

We also see that, for example,

513513513=513+13+13=51=5 

which means that 513=35,  which can be generalised to

a1n=na. 

By combining this definition with one of the previous laws of exponents ((am)n=amn) gives that for all a0 it holds that

amn=(am)1n=nam 

or

amn=(a1n)m=(na)m. 

Example 8

  1. 2713=327=3  as 333=27
  2. 1000−13=1100013=1(103)13=110331=1101=110
  3. 18=1812=1(23)12=1232=2−32
  4. 116−13=1(24)−13=12−43=2−(−43)=243


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 1 then the power is larger the larger the exponent. On the other hand, if the base lies between 0 and 1 then the power decreases as the exponent grows.

Example 9

  1. 356334  as the base 3 is greater than 1 and the first exponent 56 is greater than the second exponent 34.
  2. 3−343−56  as the base is greater than 1 and the exponents satisfy −34−56.
  3. 035034as the base 03 is between 0 and 1 and 54.

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. 532432  as the base 5 is larger than the base 4 and both powers have the same positive exponent 32.
  2. 2−533−53  as the bases satisfy 23 and the powers have a negative exponent −53.

Sometimes powers must be rewritten in order to determine the relative sizes. For example to compare 1252 with 363one can rewrite them as

1252=(53)2=56and363=(62)3=66

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 written about 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}

    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   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 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} 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,
    \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 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

Retrieved from "http://wiki.math.se/wikis/2008/bridgecourse1-ImperialCollege/index.php/1.3_Powers"