Question

When you calculate n⁰, You can type equation like this: $n^0 = \frac{n^x}{n^x} = n^{x-x} = 0$. But why $\frac{n^x}{n^x}$ is $n^{x-x}$?

Answer 1

Answer:

Step-by-step explanation:

Quick answer:

The law of exponents states that

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

Therefore

$n^{0} = {n^{x-x}} = {n^x}{n^{-x}}= \frac{n^x}{n^{-x}}$

But why is $n^{-1} = \frac{1}{n^{1}}$

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

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

Answer 2

Answer:

see explanation

Step-by-step explanation:

$\frac{n^x}{n^x}$

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

$\frac{a^b}{a^c} =a^{b-c}$

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

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

$x-x=0$

$n^{x-x}=n^0$