Evaluate: (26)0 - Mathematics

Evaluate: (2⁶)⁰

योग

(2⁶)⁰ = 2^{(6 ×} ⁰⁾ = 2⁰ = 1

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Laws of Exponents (Through Observing Patterns to Arrive at Generalisation.)

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अध्याय 5: Exponents (Including Laws of Exponents) - Exercise 5 (B)

अध्याय 5 Exponents (Including Laws of Exponents)
Exercise 5 (B) | Q 2.3

Evaluate: 8³ x 8^-5 x 8⁴

Evaluate: 4⁴ ÷ 4³ x 4⁰

Simplify, giving Solution with positive index

(-3)² (3)³

Simplify, giving Solution with positive index

(5a²b) (2ab²) (a³b)

Simplify, giving Solution with positive index

(10²)³ (x⁸)¹²

Simplify, giving Solution with positive index:

(n²)² (- n²)³

Simplify, giving Solution with positive index

- (3ab)² (-5a²bc⁴)²

Simplify and express the Solution in the positive exponent form:

`("a"^3 "b"^(-5))^-2 = "a"^(3 xx -2) "b"^(-5 xx -2)`

Evaluate: `(2^2)^0 + 2^-4 div 2^-6 + (1/2)^-3`

If m² = -2 and n = 2; find the values of: 6m^-3 + 4n²