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Calculate: 8^2/3/3√2 × 4^−5 - Mathematics

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प्रश्न

Calculate:

`8^(2/3)/(root(3)(2 xx 4^-5)`

योग
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उत्तर

Given,

`8^(2/3)/(root(3)(2 xx 4^-5)`

We need to simplify the given terms.

Thus, `8^(2/3)/(root(3)(2 xx 4^-5)`

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

⇒ `(2)^(3 xx 2/3)/((2^(1/3) xx 4^(-5 xx 1/3))`  ...[∴ (an)m = anm, an × bn = (ab)n]

⇒ `(2)^2/((2^(1/3) xx 4^(-5/3))`

⇒ `(2)^2/(2^(1/3) xx (2^2)^(- 5/3)`

⇒ `(2)^2/(2^(1/3) xx (2)^(2 xx (-5)/3)`

⇒ `(2)^2/(2^(1/3) xx (2)^((-10)/3)`  ...[∴ (an)m = anm]

⇒ `(2)^2/2^((1/3 - 10/3))`  ...[∴ an × am = an + m]

⇒ `(2)^2/2^((-9)/3) = (2)^2/2^-3`

⇒ (2)2 × 23  ...`[∴ a^n = 1/a^-n]`

⇒ 25 = 32   ...[∴ an × am = an + m]

Hence, the required is 32.

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अध्याय 6: Indices - EXERCISE 6 [पृष्ठ ६७]

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बी निर्मला शास्त्री Mathematics [English] Class 9 ICSE
अध्याय 6 Indices
EXERCISE 6 | Q 8. (v) | पृष्ठ ६७
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