Simplify the following: (i) [(64)-2/3 2-2 + 80]-1/2

(i) [(64)^-2/3 2^-2 + 8⁰]^-1/2

(ii) 3ⁿ × 9^{n + 1} ÷ (3^{n – 1} × 9^{n – 1}).

Solution:

(i) [(64)^-2/3 2^-2 + 8⁰]^-1/2

We can write it as

= [4 × 1 × 1]^-1/2

= (4)^-1/2

Here

= (2 × 2)^-1/2

= (2)^{2 × -1/2}

= (2)^-1

= 1/(2)¹

= ½

(ii) 3ⁿ × 9^{n + 1} ÷ (3^{n – 1} × 9^{n – 1})

We can write it as

= 3ⁿ × (3 × 3)^{n + 1} ÷ (3^{n – 1} × (3 × 3)^{n – 1})

By further calculation

= 3ⁿ × (3)^{2 × (n + 1)} ÷ (3^{n – 1} × (3)^2(n-1)])

= 3ⁿ × (3)^{2n + 2} ÷ (3^{n – 1} × (3)^{2n – 2})

So we get

= (3)^{n + 2n + 2} ÷ (3)^{n – 1 + 2n – 2}

= (3)^{3n + 2} ÷ (3)^{3n – 3}

Here

= (3)^{3n + 2 – 3n + 3}

= (3)⁵

We get

= 3 × 3 × 3 × 3 × 3

= 243

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