Why does $3^{16} \times 7^{-6}$ become $\frac{3^{16}} {7^{6}}$?

Question

I was doing an exercise on exponents:

$$\begin{align} \left(3^{-8} \times 7^3\right)^{-2} &= \left(3^{-8}\right)^{-2}\times \left(7^3\right)^{-2} \\ &= 3^{16} \times 7^{-6} \\ &= \frac{3^{16}} {7^{6}} \\ \end{align}$$

Why did $7^{-6}$ turn to $7^{6}$? More generally, why does a negative exponent turn positive when moved to the denominator? Would appreciate kindergarten language ;-D

Answer 1

Notice that $7^6\cdot 7^{-6}=7^{6-6}=7^0=1$

Notice also that $7^6\cdot\frac{1}{7^6}=\frac{7^6}{7^6}=1$

So, we learned that $7^6\cdot 7^{-6}=7^6\cdot\frac{1}{7^6}$.

Remembering that $x\cdot a=x\cdot b$ for nonzero $x$ implies that $a=b$ by cancelling this tells us that $7^{-6}=\frac{1}{7^6}$

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In general the following properties are useful to know:

• $x = \frac{x}{1}=x^1$
• $x^n = \frac{1}{x^{-n}}$
• $x^{-n}=\frac{1}{x^n}$

Another useful identity is $x^0 = 1$ which is true for all nonzero $x$

Tangentially, depending on context it can also be correct to say that $x^0=1$ for $x=0$ as well, for example in the field of combinatorics. There are some other situations though where we leave $0^0$ undefined.

Answer 2

First you need to understand why $$ 7^{-1} = \frac{1}{7} $$ In the row below, to move one to the right, multiply by $7$: $$ 7^1=7,\qquad 7^2=49,\qquad 7^3=343,\qquad\dots $$ And consequently, to move to the left, divide by $7$. So that is how to extend it the other way: keep dividing by $7$: $$ \dots \qquad7^{-2} = \frac{1}{49},\qquad 7^{-1} = \frac{1}{7},\qquad7^0=1,\qquad7^1=7,\qquad 7^2=49,\qquad\dots $$

Answer 3

Exponentiation is initially understood as repeated multiplication with the exponent indicating how many times the base appears as a factor in the product. Thus, for example, $2^3=2\times2\times2$ since $2$ appears as a factor three times.

Given this initial understanding of exponentiation certain properties can be observed such as $x^2\cdot x^3=(x\cdot x)\cdot(x\cdot x\cdot x)=x^5$. From examples such as this we discover that

$$ x^m\cdot x^n=x^{m+n} \tag{1}$$

Similarly, it can be seen that $(x^3)^2=x^3\cdot x^3=x^6$, for example. From such examples we discover that

$$(x^m)^n=x^{mn} \tag{2}$$

But these rules assume that $n$ is a positive integer. What if that is not the case? Can meaning be given to exponentiation in such a way that rules $(1)$ and $(2)$ are preserved?

Let us see, for example, if meaning can be given to exponentiation with $0$.

We would want it to be true that $x^0\cdot x^n=x^{0+n}=x^n$ in order to satisfy rule ${1}$. But that means that it would have to be the case that $x^0=1$. So that is how exponentiation to the power $0$ is defined.

But then what about exponentiation to a negative integer power? How could sense be given to a term like $x^{-n}$?

Well, in order for rule $(1)$ to apply it would have to be the case that

$$ x^{-n}\cdot x^n=x^{-n+n}=x^0=1 $$

But if $x^{-n}\cdot x^n=1$ then it must be the case that $x^{-n}=\dfrac{1}{x^n}$ and $x^n=\dfrac{1}{x^{-n}}$.

Answer 4

The OP asked for kindergarten language. At that grade level math rules are stressed. So if you are taking algebra-precalculus and this is your last math course, here are a couple of rules;

$\tag 1 \frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d} \;\text{ where } b \ne 0 \text{, } d \ne 0$

$\tag 2 x = \frac{x}{1}$

$\tag 3 x^m x^n = x^{m+n} \;\text{ where } x \gt 0 \text{ and both } m \text{ and } n \text{ are integers}$

$\tag 4 x \times 1 = x$

$\tag 5 \frac{x}{x} = 1 \;\text{ where } x \ne 0$

$\tag 6 x^0 = 1 \;\text{ where } x \gt 0$

$\tag 7 (-x) + (+x) = 0 \;\text{ for any } x$

So if for some reason your instructor wants you to get rid of negative exponents, and you see $3^{16} \times 7^{-6}$, you can use these seven rules:

$3^{16} \times 7^{-6} = \frac{3^{16} \times 7^{-6}}{1} \; \text{ by (2)}$

$\frac{3^{16} \times 7^{-6}}{1} = \frac{3^{16} \times 7^{-6}}{1} \times 1 \; \text{ by (4)}$

$\frac{3^{16} \times 7^{-6}}{1} \times 1 = \frac{3^{16} \times 7^{-6}}{1} \times \frac{7^6}{7^6} \; \text{ by (5)}$

$\frac{3^{16} \times 7^{-6}}{1} \times \frac{7^6}{7^6} = \frac{3^{16} \times 7^{-6} \times 7^{+6}} {1 \times 7^{+6} } \; \text{ by (1)}$

$\frac{3^{16} \times (7^{-6} \times 7^{+6})} {1 \times 7^{+6} } = \frac{3^{16} \times (7^{-6 + +6}) } {7^{+6} } \; \text{ by (3) & (4)}$

$\frac{3^{16} \times (7^{(-6) + (+6)}) } {7^{+6} } = \frac{3^{16} \times (7^{0})} {7^{+6}} = \frac{3^{16} \times 1} {7^{+6}} \; \text{ by (7) & (6)}$

$\frac{3^{16} \times 1} {7^{+6}} = \frac{3^{16}} {7^{6}}\; \text{ by (4)}$

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You now have a new rule that you can use:

$\tag 8 x \times y^{-n} = \frac{x}{y^n} \;\text{ where } y \gt 0 \text{ and } n \text{ is any integer}$

Answer 5

$$ 7^{-6}= 7^{6*-1} = {(7^6)}^{-1}$$ and $${x}^{-1}=\frac{1}{x}$$ so $$ 7^{-6}= {(7^6)}^{-1}=\frac{1}{{7}^{6}}$$ the first and second equation could be proven using the laws of exponents.

Answer 6

Short answer: Because $7^{-6}$ is defined as $1/7^6$. We cannot argue about the truth of definitions!

The better question would be "why it was defined this way". And this was mostly be done because we have the rule

$$a^n\times a^m=\underbrace{a\times\cdots\times a}_n\times\overbrace{a\times\cdots\times a}^m=\underbrace{a\times\cdots\times a}_{n+m}=a^{n+m}$$

for $n,m=1,2,3,4,...$, and we want to keep this very nice rule for other exponents. But if we really want to keep it, above definition is forced onto us:

$$7^{-6}\times7^6=7^{-6+6}=7^0=1\qquad\implies\qquad7^{-6}=1/7^6.$$

You can go on and ask why $7^0=1$ though.

Answer 7

The question has thoroughly been answered but I do want to contribute with this table that I like.

Begin with this table:

\begin{array}{ccl} 7^5 & = & 7\cdot 7\cdot 7\cdot 7\cdot 7 \\ 7^4 & = & 7\cdot 7\cdot 7\cdot 7 \\ 7^3 & = & 7\cdot 7\cdot 7 \\ 7^2 & = & 7\cdot 7 \\ 7^1 & = & 7 \end{array} Going up:

• The previous line is multiplied by $7$ on the right hand side.
• The exponent is increased by $1$ on the left hand side.

Going down:

• The previous line is divided by $7$ on the right hand side.
• The exponent is decreased by $1$ on the left hand side.

What would be a natural extension? To keep doing this pattern both up and down! And thus:

\begin{array}{ccl} 7^5 & = & 7\cdot 7\cdot 7\cdot 7\cdot 7 \phantom{\dfrac1{7^3}} \\ 7^4 & = & 7\cdot 7\cdot 7\cdot 7\phantom{\dfrac1{7^3}} \\ 7^3 & = & 7\cdot 7\cdot 7\phantom{\dfrac1{7^3}} \\ 7^2 & = & 7\cdot 7 \phantom{\dfrac1{7^3}} \\ 7^1 & = & 7\phantom{\dfrac1{7^3}} \\ 7^0 & = & \dfrac77 = 1\phantom{\dfrac1{7^3}} \\ 7^{-1} & = & \dfrac17 \phantom{\dfrac1{7^3}} \\ 7^{-2} & = & \dfrac{1/7}7 = \dfrac1{7^2} \\ 7^{-3} & = & \dfrac{1/7^2}7 = \dfrac1{7^3} \end{array}