Q. 723.6( 10 Votes )

# Find the value of:

(a) 7^{0}

(b) 7^{7} ÷ 7^{7}

(c) (–7)^{2 × 7 – 6 – 8}

(d) (2^{0} + 3^{0} + 4^{0}) (4^{0} – 3^{0} – 2^{0})

(e) 2 × 3 × 4 ÷ 2^{0} × 3^{0} × 4^{0}

(f) (8^{0} – 2^{0}) × (8^{0} + 2^{0})

Answer :

**(a)** We have the law of Indices as follows:

a^{0} = 1

Where a = any integer

7^{0} = 1

**(b)** According to Law of indices,

a^{x} ÷ a^{y} = a ^{(x - y)}

Here in the question,

a = 7

x = 7

y = 7

∴ x - y = 7 - 7

= 0

7^{7} ÷ 7^{7} = 7^{0} = 1

**(c)** (–7)^{2 × 7 – 6 – 8}

= (–7)^{14-14}

= (–7)^{0}

= 1

**(d)** (2^{0} + 3^{0} + 4^{0}) (4^{0} – 3^{0} – 2^{0}) = (1 + 1 + 1) × (1-1-1)

= 3 × -1

= -3

**(e)** Let us evaluate the terms separately,

2 × 3 × 4 = 6 × 4

= 24

2^{0} × 3^{0} × 4^{0} = 1 × 1 × 1

= 1

2 × 3 × 4 ÷ 2^{0} × 3^{0} × 4^{0} = 24 / 1

= 24

**(f)** (8^{0} – 2^{0}) × (8^{0} + 2^{0})

= (1 – 1) × (1 + 1)

= 0 × 1

= 0

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