Powers with natural exponents

An expression of the form

3^7

(say: "three raised to the power of seven", or "three to the power of seven") or more generally

a^n

is called a power. The $3$ or the $a$ is called the base, and the $7$ or the $n$ is called the exponent. Both the base and the exponent can be real numbers, but here we focus on $n$ being a natural number or $0$ , that is, we assume that $n\in \{0,1,2,3,...\}$ . In later chapter we will discuss powers where the exponent $n$ can be negative, or a fraction.

By definition, the exponent tells you how often the bases has to be multiplied with itself. So for the power $3^7$ we have to multiply $3$ seven times with itself:

3^7 =\underbrace{3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3}_{7}

More generally we define for any natural number $n$ and any real number $a$ :

\boxed{a^n =\underbrace{a\cdot ... \cdot a}_{n}}

We also define

\boxed{a^0= 1}

To see that this last definition actually makes sense, have a look at the following powers:

\begin{array}{lll} 2^3&=8 \quad\quad :2\\ 2^2&=4 \quad\quad :2\\ 2^1&=2 \quad\quad :2\\ 2^0&=\, ? \end{array}

Notice that decreases the exponent by $1$ divides the power by $2$ . So if we want to maintain this pattern (and we want to do this), it is clear that $2^0=1$ . And more generally, $a^0=1$ .

If a bracket is raised to some power, then we have to multiply everything in the brackets with itself, e.g.

(ab)^3=\underbrace{(ab)\cdot (ab)\cdot (ab)}_{3} = ab ab ab

(a+b)^3=\underbrace{(a+b)\cdot (a+b)\cdot (a+b)}_{3}

Note that we evaluate powers before we evaluate anything else, e.g.:

2+4^3=2+4\cdot 4\cdot 4 \text{, so } 2+4^3\neq 6^3

2\cdot 4^3=2\cdot 4\cdot 4\cdot 4 \text{, so } 2\cdot 4^3\neq 8^3

2a^3=2\cdot a\cdot a\cdot a \text{, so } 2\cdot a^3\neq (2a)^3

-2^4=-2\cdot 2\cdot 2\cdot 2 = -16 \text{, so } -2^4\neq 16

Recall that a similar rule exists for multiplication. Multiplication is evaluated before addition. So,

2+4\cdot 4\cdot 4 = 2+64 \text{, so } 2+4\cdot 4\cdot 4 \neq 6\cdot 4\cdot 4

There are some nice rules that tell you how to deal with products of powers, the so called power rules. But before we discuss them, solve some exercises.

Exercise 1

Are the statements correct, and why? Argue with the definition.

$3+5\cdot 2 = 30$
$3\cdot 4^2 =144$
$0^7=0$ and $a^1=a$
$3^4=81$
$10^4=10000$
$a^3 \cdot a^5 = a^8$
$a^3 \cdot a^5 \cdot a^2 = a^9$
$(a+b)^4=a^4+b^4$
$(a\cdot b)^4=a^4 \cdot b^4$
$(a^3)^5=a^{15}$
$\frac{a^8}{a^4}=a^{2}$
$\frac{a^7}{a^3}=a^{4}$
$a^5\cdot b^5=(a\cdot b)^5$
$a^5\cdot b^4= (ab)^9$
$a^5\cdot b^4= (ab)^{20}$
$\frac{a^5}{a^7}=\frac{1}{a^2}$
$(2a)^4=16a^4$
$a^5\cdot b^5=a b^5$
$2a^4=16 a^4$
$\frac{a^4\cdot b^7\cdot c^2}{b^4\cdot a^5\cdot c^4}=\frac{b^3}{a\cdot c^2}$
$(a+b)^5\cdot (a+b)^4=(a+b)^9$
$\frac{a^5}{a^4}=a$
$\frac{(a+b)^3}{(a+b)^4}=\frac{1}{a+b}$
$(2a)^3 \cdot (3a)^2=72a^5$
$\frac{a^5}{3a^2}=3 a^3$
$a+b^3=(a+b)^3$
$\left(\frac{1}{a}\right)^4=\frac{1}{a^4}$
$\left(\frac{2}{a}\right)^4=\frac{16}{a^4}$
$\frac{2a^2}{a^2}=2$
$\frac{2+a^2}{a^2}=2$

Solution

$3+5\cdot 2 = 3+5+5 = 13 \neq 30$ wrong
$3\cdot 4^2 = 3\cdot 4\cdot 4 \neq 144$ wrong
$0^7=0\cdot 0\cdot 0\cdot 0\cdot 0\cdot 0\cdot 0=0$ correct, and $a^1=a$ correct
$3^4=3\cdot 3\cdot 3\cdot 3 =81$ correct
$10^4=10 \cdot 10\cdot 10\cdot 10 =10000$ correct
$a^3 \cdot a^5 = aaa\,aaaaa = a^8$ correct
$a^3 \cdot a^5 \cdot a^2 = aaa\,aaaaa\,aa =a^{10} \neq a^9$ wrong
$(a+b)^4=(a+b)(a+b)(a+b)(a+b) \neq a^4+b^4$ wrong
$(a\cdot b)^4= (ab)(ab)(ab)(ab)=abababab=aaaabbbb=a^4 \cdot b^4$ correct
$(a^3)^5=(a^3)(a^3)(a^3)(a^3)(a^3) = aaa\,aaa\,aaa\,aaa\,aaa = a^{15}$ correct
$\frac{a^8}{a^4}=\frac{aaaa\not a \not a \not a \not a}{\not a \not a \not a \not a} = a^4 \neq a^{2}$ wrong
$\frac{a^7}{a^3} = \frac{aaaa\not a \not a \not a}{\not a \not a \not a} = a^{4}$ correct
$a^5\cdot b^5= aaaaa\,bbbbb = ababababab = (ab)(ab)(ab)(ab)(ab) = (a\cdot b)^5$ correct
$a^5\cdot b^4= aaaaa\,bbbb \neq (ab)^9$ wrong
$a^5\cdot b^4= aaaaa\,bbbb \neq (ab)^{20}$ wrong
$\frac{a^5}{a^7} = \frac{\not a \not a \not a \not a \not a}{\not a \not a \not a \not a \not a aa} = \frac{1}{a^2}$ correct
$(2a)^4 = (2a)(2a)(2a)(2a) = 2\cdot 2 \cdot 2 \cdot 2 \cdot aaaa = 16a^4$ correct
$a^5\cdot b^5 = aaaaa\,bbbbb = ababababab=(ab)^5 \neq a b^5$ wrong
$2a^4= 2\cdot aaaa \neq 16 a^4$ wrong
$\frac{a^4\cdot b^7\cdot c^2}{b^4\cdot a^5\cdot c^4}= \frac{ \not a \not a \not a \not a \not b \not b \not b \not b bbb \not c \not c}{\not b \not b \not b \not b a \not a \not a\not a \not a \not c \not c cc} = \frac{b^3}{a\cdot c^2}$ correct
$(a+b)^5\cdot (a+b)^4 = (a+b)(a+b)(a+b)(a+b)\,\,(a+b)(a+b)(a+b)(a+b)(a+b) = (a+b)^9$ correct
$\frac{a^5}{a^4}= = \frac{a \not a \not a \not a \not a}{ \not a \not a \not a \not a} = a$ correct
$\frac{(a+b)^3}{(a+b)^4} = \frac{\cancel{(a+b)}\cancel{(a+b)}\cancel{(a+b)}}{\cancel{(a+b)} \cancel{(a+b)} \cancel{(a+b)} (a+b)} = \frac{1}{a+b}$ correct
$(2a)^3 \cdot (3a)^2 = (2a)(2a)(2a)\,(3a)(3a)=2\cdot 2\cdot 2\cdot 3\cdot 3\cdot aaaaa = 72 a^5$ correct
$\frac{a^5}{3a^2} = \frac{aaa \not a \not a}{3 \not a \not a} = \frac{a^3}{3} \neq 3 a^3$ wrong
$a+b^3 = a+bbb \neq (a+b)^3$ wrong
$\left(\frac{1}{a}\right)^4 = \frac{1}{a}\frac{1}{a}\frac{1}{a}\frac{1}{a} = \frac{1\cdot 1\cdot 1\cdot 1}{aaaa} = \frac{1}{a^4}$ correct
$\left(\frac{2}{a}\right)^4= \frac{2}{a}\frac{2}{a}\frac{2}{a}\frac{2}{a} = \frac{2\cdot 2\cdot 2\cdot 2}{aaaa} = \frac{16}{a^4}$ correct
$\frac{2a^2}{a^2}= \frac{2 \not a \not a}{\not a \not a} =2$ correct
$\frac{2+a^2}{a^2}= \frac{2+aa}{aa} \neq 2$ wrong

Exercise 2

Determine $(-1)^3$ , $(-1)^5$ , $(-1)^2$ , and $(-1)^4$ .
Formulate a rule that tells you if $(-1)^n$ is $1$ or $-1$ based on the value of $n$ (a natural number).
Is $(-2)^{1001}$ positive or negative? Why?
What is the difference between $(-2)^{1000}$ and $-2^{1000}$ ?

Solution

$(-1)^3=(-1)(-1)(-1)=\underline{-1}$ , $(-1)^5=(-1)(-1)(-1)(-1)(-1)=\underline{-1}$ , $(-1)^2=(-1)(-1)=\underline{1}$ $(-1)^4=(-1)(-1)(-1)(-1)=\underline{1}$
$(-1)^n=-1$ for $n$ odd, and $(-1)^n=1$ for $n$ even.
negative, because $n$ is odd: $(-2)^{1001}=\underbrace{(-2)....(-2)}_{1001}<0$ .
$(-2)^{1000}=\underbrace{(-2)....(-2)}_{1000}>0$ is positive, $-2^{1000}=-\underbrace{\,2\cdot 2\cdot ...\cdot 2}_{1000}<0$ is negative.

Exercise 3

The power rules. $a$ and $b$ are some numbers, $n$ and $m$ are natural numbers. Fill in the boxes.

(SB) Same base rules: $a^n \cdot a^m = a^\square, \quad$ $\frac{a^n}{a^m} = a^\square,\quad$ $\frac{a^n}{a^m} = \frac{1}{a^\square}$
(SE) Same exponent rules: $a^n \cdot b^n = (a\cdot b)^\square, \quad$ $\frac{a^n}{b^n}=\left(\frac{a}{b}\right)^\square$
(PP) Power of a power rule: $\left(a^n\right)^m = a^\square$
(DF) Do not forget rule: $a^\square + b^\square \neq (a+b)^\square$

Solution

(SB) Same base rules:

$a^n \cdot a^m = a^{n+m}$

$\frac{a^n}{a^m} = a^{n-m}\,$ if $n>m$

$\frac{a^n}{a^m} = \frac{1}{a^{m-n}}\,$ if $n<m$
(SE) Same exponent rules:

$a^n \cdot b^n = (a\cdot b)^n$ ,

$\frac{a^n}{b^n}=\left(\frac{a}{b}\right)^n$
(PP) Power of a power rule: $\left(a^n\right)^m = a^{n\cdot m}$
(DF) Do not forget rule: $a^n + b^n \neq (a+b)^n$