Question

When you calculate n⁰, You can type equation like this:
`n^0 = (n^x)/(n^x) = n^(x-x) = 0`. But why
`(n^x)/(n^x)` is
`n^(x-x)`?

Answer 1

Explanation:

Quick answer:

The law of exponents states that

`n^(-1) = (1)/(n^(1))`

Therefore

`n^(0) = {n^(x-x)} = {n^x}{n^(-x)}= (n^x)/(n^(-x))`

But why is
`n^(-1) = (1)/(n^(1))`

This is because

`n^(x+1) = n^x * n^1`

and inversely, since division is the inverse of multiplication

`n^(x-1) = n^x / n^1`

Putting x=0 gives

`n^(-1) = (1)/(n^(1))`

Answer 2

see explanation

Explanation:

`(n^x)/(n^x)`

In exponent division rule, you subtract the exponents when dividing numbers with the same base.

`(a^b)/(a^c) =a^(b-c)`

The bases
`n` are same, when dividing, subtract the exponents
`x`.

`(n^x)/(n^x) =n^(x-x)`

`x-x=0`

`n^(x-x)=n^0`