Question

Prove that,
$\sf {x}^{0} = 1 \: \: \: \: (x \neq 0)$
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Answer 1

Answer:

$\rm{Prove \: that \: x⁰=1} \\ \rm \: RHS=1 \\ \: \: \: \: \: \: \: = \frac{1}{1} \\ \rm \: {Multiply \: numerator \: denominator \: by} \\ \rm \: \: \: \: = \frac{x}{x} \\ \\ \rm= \frac{x¹}{x¹} \\ \\ \rm = {x}^{1 - 1} \\ \rm = {x}^{0} \\ \rm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rm\: \: Law \: of \: exponent= \frac{ {x}^{a} }{ {x}^{b} } = {x}^{a - b} \\ \\ \\ \huge\rm{Hence \: proved!}$

Answer 2

Solution :-

We need to prove that x⁰ = 0 . Taking LHS and simplifying .

As we know that , $\tt a^0$ can be written as $\tt a^{1-1}$ . Similarly ,

$\displaystyle \longrightarrow\tt x^{1-1}$

Using laws of exponents $\; \sf\Big[x^{m-n} = \dfrac{x^m}{x^n}\Big]$

$\\ \displaystyle\longrightarrow \tt \dfrac{x^1}{x^1}=\dfrac{x}{x}\\\\\displaystyle\longrightarrow \boxed{\red{\tt{1}}}$

Now , comparing with RHS

• LHS = RHS

Hence proved !

More information :-

$\begin{gathered}\boxed{\begin{minipage}{5 cm}\bf{\dag}\:\:\underline{\text{Law of Exponents :}}\\\\\bigstar\:\:\sf\dfrac{a^m}{a^n} = a^{m - n}\\\\\bigstar\:\:\sf{(a^m)^n = a^{mn}}\\\\\bigstar\:\:\sf(a^m)(a^n) = a^{m + n}\\\\\bigstar\:\:\sf\dfrac{1}{a^n} = a^{-n}\\\\\bigstar\:\:\sf\sqrt[\sf n]{\sf a} = (a)^{\dfrac{1}{n}}\end{minipage}}\end{gathered}$