Question

Prove that ${x}^{0} = 1$
Where $x$is not equal to 0.

Answer 1

$\huge{\bold{Hey \:Mate}}$

$\large{\bold{Solution:}}$

As we can know the rule

↓ ↓ ↓ ↓ ↓ ↓

$\huge{\fbox{\bold{x^{a} \times x ^{b} = x^{a + b} }}}$

So, we can multiply $\bold{x ^{1} }$ to both the sides, That is :—

= > $\large{\fbox{\bold{x ^{0} \times x^{1} = 1 \times x^{1} }}}$

= > $\large{\fbox{\bold{x ^{1} = x ^{1} }}}$

= > $\large{\fbox{\bold{1 = 1}}}$

This signifies that the value of $\bold{x ^{0} }$
is $\bold{1}$ because according to additive identity property of multiplication$\bold{x ^{1} = x }$

So, the value of $\bold{x ^{0} }$ has to be $\textbf{1}$

Hence, proved ✔️✔️

$\large{\bold{Hope\:it\:works !!!\::)}}$