Question

why the value of
${x}^{0}$
equal to (0)​

Answer 1

Hey the question has some error

You should know that anything to the power 0 is 1

So, x^0=1

For any of the case

it is always 1:

For n a non-negative integer, we define x^n recursively:

x^0 = 1;

for x > 0, x^n = x * x^(n - 1) .

For negative integers, provided x ≠ 0 ,

we can work backwards: x^n = x^(n + 1) / x ;

or we can use x^(-n) = 1 / x^n .

These two can easily be shown to be equivalent.

Now for the index being a member of the set of reals:

For x > 0 , x^0 = 1 .

For x = 0 , a limit argument (invoking logs) would make it 1, also.

(For x < 0 , we start to need complex numbers.)

Shortly, x^0 is usually 1.

Hope it helps

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