Question

why anything to the power 0 always 1​

Answer 1

Solution 1):-

→ a^x = a^x

when we add zero in any number result is same . so, x can we written as (x + 0),

→ a^x = a^(x + 0)

using a^(m + n) = a^m * a^n

→ a^x = a^x * a^0

a^x will be cancel from both sides,

→ 1 = a^0

therefore, power of any number zero is always equal to 1 .

Solution 2) :-

→ a^x * a^(-x) = a^x * a^(-x)

using a^(-m) = 1/a^m in LHS,

→ a^x * (1/a^x) = a^x * a^(-x)

→ 1 = a^x * a^(-x)

now using a^m * a^n = a^(m + n) in RHS,

→ 1 = a^(x - x)

→ 1 = a^0

therefore, power of any number zero is always equal to 1 .

Solution 3) :-

we know that,

→ a^(x - 1) = a^x ÷ a

putting x = 1,

→ a^(1 - 1) = a^1 ÷ a

→ a^0 = a ÷ a

→ a^0 = 1

therefore, power of any number zero is always equal to 1 .

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Answer 2

SOLUTION

TO EXPLAIN

why anything to the power 0 always 1

FORMULA TO BE IMPLEMENTED

We are aware of the formula on indices that :

‌

‌$\sf{1. \: \: {a}^{m} \times {a}^{n} = {a}^{m + n} }$

‌$\displaystyle \sf{2. \: \: \: \frac{ {a}^{m} }{ {a}^{n} } = {a}^{m - n} }$

‌$\displaystyle \sf{3. \: \: \: { ({a}^{m} )}^{n} = {a}^{mn} }$

‌

EVALUATION

Let a be the number then we have to prove that

$\displaystyle \sf{ \: \: {a}^{0} = 1}$

Now

$\displaystyle \sf{ {a}^{0} }$

$\displaystyle \sf{ = {a}^{m - m} }$

$\displaystyle \sf{ = \frac{{a}^{m }}{{a}^{m }} } \: \: \:( by \: formula \: 2)$

$\displaystyle \sf{ = 1}$

Hence proved

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