Question

@Osako

Answer 1

if you write 1 as x^0 and subtract exponents (top exponent minus bottom exponent) what do you get ?

Answer 2

in other words simplfy
\[ \frac{ x^0}{x^{-1} }\]
using the rule "subtract exponents" like you did on the earlier problem

Answer 3

I'm confused

Answer 4

0-(-1)?

Answer 5

yes

Answer 6

so -1 is the answer? why can you just get rid of the 1 for a 0

Answer 7

?
are you saying 0 - (-1) is -1 ?

Answer 8

1*

Answer 9

sorry, lol

Answer 10

so 1 is the exponent. the base is x. the answer is x^1 or just x

so notice
\[ \frac{1}{x^{-1}} = x^1 \]

Answer 11

notice this:

you know x divided by itself is 1
\[ \frac{x}{x} = 1\]
If we use the "subtract exponents" rule on

\[ \frac{x^1}{x^1} = \]
you get ?

Answer 12

you get an exponent of 1-1 = 0 and a base of x
you get
\[ \frac{x}{x}= x^0 \]
but x/x is 1, so we find
\[ 1 = x^0\]

Answer 13

0?

Answer 14

thank you

Answer 15

we can use the rule to show
\[ \frac{1}{x} = \frac{x^0}{x^1} \]
now use the subtract rule to get
\[ \frac{x^0}{x^1} = x^{0-1} = x^{-1} \]
or
\[ \frac{1}{x} = x^{-1} \]

Answer 16

Could it equal 1?

Answer 17

It should based on the question I have

Answer 18

Could it equal 1 ? what is "it" ?

Answer 19

The answer to the problem

Answer 20

\[ \frac{1}{x^{-1}} = x \]

that is true for any x (except x being 0)

for example
\[ \frac{1}{2^{-1}} = 2 \]

a negative exponent means you can "flip" the number, and make the exponent positive.