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प्रश्न
Simplify:
`(((-2)/3)^-2)^3 xx (1/3)^-4 xx 3^-1 xx 1/6`
उत्तर
Using laws of exponents, (am)n = (a)m × n, `a^-m = 1/a^m`, am × an = am + n and am ÷ an = am – n ...[∵ a is non-zero integer]
∴ `(((-2)/3)^-2)^3 xx (1/3)^-4 xx 3^-1 xx 1/6 = ((-2)/3)^((-2) xx 3) xx (3)^4 xx 1/3 xx 1/6`
= `((-2)/3)^-6 xx 3^4 xx 1/3 xx 1/(2 xx 3)` ...[∵ 6 = 2 × 3]
= `(3/(-2))^6 xx 3^4 xx 1/(3 xx 2 xx 3) `
= `(3)^6/(-2)^6 xx 3^4 xx 1/(2^1 xx 3^2)`
= `(3)^(6 + 4)/((2)^(6 + 1) xx 3^2)` ...[(–am) = am, If m is an even number]
= `(3)^10/(2^7 xx 3^2)`
= `3^(10 - 2)/2^7`
= `3^8/2^7`
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