Evaluate: (26)0

Evaluate: (2⁶)⁰

बेरीज

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

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

Evaluate: (3⁰)⁶

Evaluate: 5⁴ × 5³ ÷ 5⁵

Evaluate: 5⁴ ÷ 5³ x 5⁵

Evaluate: 4⁴ ÷ 4³ x 4⁰

Simplify, giving Solution with positive index

`2^"4a". 2^("3a") .2^(-"a")`

Simplify, giving Solution with positive index

(10²)³ (x⁸)¹²

Simplify, giving Solution with positive index:

(n²)² (- n²)³

Simplify, giving a Solution with a positive index.

`(1/"2x")^3 xx (6"x")^2`

Simplify, giving Solution with positive index

`((5"x"^7)^3 . (10"x"^2)^2)/(2"x"^6)^7 = (5^3 "x"^(7xx3) . 10^2 . "x"^(2xx2))/(2^7. "x"^(6xx7))`

Simplify and express the Solution in the positive exponent form:

`((2^3)^5 xx 5^4)/(4^3 xx 5^2)`