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

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

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.

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