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Find the value of 81^3/4 − (1/32)^−2/5 + 8^1/3 ⁢(1/2)^−1 ⁢(4)^0 - Mathematics

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Question

Find the value of `81^(3/4) - (1/32)^((-2)/5) + 8^(1/3) (1/2)^-1 (4)^0`

Sum
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Solution

Given,

`81^(3/4) - (1/32)^((-2)/5) + 8^(1/3) (1/2)^-1 (4)^0`

We need to simplify the given terms.

Thus, `81^(3/4) - (1/32)^((-2)/5) + 8^(1/3) (1/2)^-1 (4)^0`

⇒ `(3^4)^(3/4) - (1/2^5)^((-2)/5) + (2^3)^(1/3) xx (1/2)^-1 xx 1`  ...[∴ a0 = 1 where a ≠ 0]

⇒ `(3)^(4 xx 3/4) - (2^-5)^((-2)/5) + (2)^(3 xx 1/3) xx (2^-1)^-1`  ...`[∴ a^-n = 1/a^n, (a^m)^n = a^(mn)]`

⇒ (3)3 – (2)2 + (2) × 2

⇒ 27 – 4 + 4

⇒ 27

Hence, `81^(3/4) - (1/32)^((-2)/5) + 8^(1/3) (1/2)^-1 (4)^0 = 27`.

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Chapter 6: Indices - EXERCISE 6 [Page 66]

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B Nirmala Shastry Mathematics [English] Class 9 ICSE
Chapter 6 Indices
EXERCISE 6 | Q 3. (v) | Page 66
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