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

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

• The convention about powers is that $1/x = x^{-1}$. That way when you multiply (say) $x^3 \times x^{-1}$ you add the exponents: $x^3 \times x^{-1}= x^{3-1}=x^2$ and so on. Commented Nov 3, 2017 at 0:49
• What do you think that $7^{-6}$ means to begin with? This should be your first question. And it should have been your question already in August. This is not to criticise, but to help: Maths is half as bad when you have an understanding of what the symbols that you manipulate mean. Commented Nov 3, 2017 at 8:16
• By dividing by 7 you decrement the exponent by 1. So what happens when you divide when you have $1 = 7^{0}$ Commented Nov 4, 2017 at 1:11
• Everyone, I'm sorry to write this late, I really appreciate all the great input I've got :). Carsten, I'm doing Algebra right now, moving from section to section and this part of solving exponents didn't come up back then. Thank you :) Commented Nov 7, 2017 at 5:59
• Carsten.. I just clicked on that question from August. Yes, I've totally forgotten about it and didn't look it up back then (skipped it). You are right, it's my fault. I probably didn't understand the concept and decided to skip it for a while. All understood now :), danke. Commented Nov 7, 2017 at 6:17

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 $7^{-6}=\frac{1}{7^6}$

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.

• This is the reason why negative exponential is defined this way: to maintain the properties of regular exponentiation with positive numbers. The matter of "negative turn positive when moved to the denominator" is simply one: its definition. Why this definition is used is another matter. Commented Nov 3, 2017 at 14:29

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 $7$: $$\dots \qquad7^{-2} = \frac{1}{49},\qquad 7^{-1} = \frac{1}{7},\qquad7^0=1,\qquad7^1=7,\qquad 7^2=49,\qquad\dots$$

• I think this would look like a clearer pattern if you were to show it vertically stacked, rather than horizontally. That way, you can compare what is happening on both sides much easier Commented Nov 3, 2017 at 0:55
• the last row is where my mind goes... except in reverse... 7^3, 7^2, ..., 7^-3, 7^-4, ...: take the pattern and go out in the different directions and see where it leads. Commented Nov 3, 2017 at 15:04
• This is the "kindergarten language" explanation that immediately comes to my mind as well – not a proof to convince you that it's true, but a mental picture to give you the right intuition. Commented Nov 4, 2017 at 0:59
• Not just kindergarten, but those at any stage prior to algebra. In case (as I suspect) the OP doesn't know algebra. Commented Nov 4, 2017 at 2:45

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 $x^n=\dfrac{1}{x^{-n}}$.

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.

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:

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

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

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

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

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

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

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

You now have a new rule that you can use:

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

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}

$$7^{-6}= 7^{6*-1} = {(7^6)}^{-1}$$ 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.