Question

Find the value of.

(i) (3⁰ + 4⁻¹) × 2² (ii) (2⁻¹ × 4⁻¹) ÷2⁻²

(iii) (iv) (3⁻¹ + 4⁻¹ + 5⁻¹)⁰

(v)

Answer 1

(i)

(ii) (2⁻¹ × 4⁻¹) ÷ 2^{− 2} = [2⁻¹ × {(2)²}^{− 1}] ÷ 2^{− 2}

= (2^{− 1} × 2^{− 2}) ÷ 2^{− 2}

= 2^{−1+ (−2)} ÷ 2⁻² (a^m × aⁿ = a^{m + n})

= 2⁻³ ÷ 2⁻²

= 2^{−3 − (−2)} (a^m ÷ aⁿ = a^{m − n})

= 2^{−3 + 2} = 2 ⁻¹

(iii)

(iv) (3⁻¹ + 4⁻¹ + 5⁻¹)⁰

= 1 (a⁰ = 1)

(v)