Warning: jsMath requires JavaScript to process the mathematics on this page.
If your browser supports JavaScript, be sure it is enabled.

1.3 Powers

From Förberedande kurs i matematik 1

(Difference between revisions)

Jump to: navigation, search

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

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

Example 2

\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}
\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

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

Example 4

\displaystyle \frac{3^{100}}{3^{98}} = 3^{100-98} = 3^2
\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

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

\displaystyle (-1)^{56} = 1\quad as \displaystyle 56 is an even number
\displaystyle \frac{1}{(-1)^{11}} = \frac{1}{-1} = -1\quad because 11 is an odd number
\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

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
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
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

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

\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.
\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.
\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

\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.
\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:

\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.

\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.

\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.

\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

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

3. Roots and logarithms

Search