English, asked by umedkhan0044, 5 months ago

8. The solution for (125)^2n/3. X (27)^-
n/6/ (75)^-n/6​

Answers

Answered by RvChaudharY50
2

Given :- The solution for (125)^2n/3 * (27)^-n/6 / (75)^-n/6 ?

Solution :-

→ { (125)^(2n/3) * (27)^(-n/6) } / (75)^(-n/6)

→ { (5³)^(2n/3) * (3³)^(-n/6) } / (75)^(-n/6)

using (a^m)^n = (a)^(m * n) in numerator now,

→ [5^{3 * (2n/3)} * 3^{3 * (-n/6)}] / (75)^(-n/6)

→ [5^(2n) * 3^(-n/2)] / (75)^(-n/6)

now, breaking the denominator part ,

→ [5^(2n) * 3^(-n/2)] / (25 * 3)^(-n/6)

→ [5^(2n) * 3^(-n/2)] / (5² * 3)^(-n/6)

using (a * b)^m = a^m * b^m in denominator, we get,

→ [5^(2n) * 3^(-n/2)] / [(5²)^(-n/6) * 3^(-n/6)]

→ [5^(2n) * 3^(-n/2)] / [5^{2 * (-n/6)} * 3^(-n/6)]

→ [5^(2n) * 3^(-n/2)] / [5^(-n/3) * 3^(-n/6)]

now, using a^m / a^n = a^(m - n) in both terms we get,

→ 5^{2n - (-n/3)} * 3^{(-n/2) - (-n/6)}

→ 5^{(6n + n)/3} * 3^{(-3n + n)/6}

→ 5^(7n/3) * 3^(-2n/6)

→ 5^(7n/3) * 3^(-n/3)

finally using a^(-m) = 1/a^m , we get,

[ 5^(7n/3) / 3^(n/3) ] (Ans.)

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Answered by Anonymous
3

Given :- The solution for (125)^2n/3 * (27)^-n/6 / (75)^-n/6 ?

Solution :-

→ { (125)^(2n/3) * (27)^(-n/6) } / (75)^(-n/6)

→ { (5³)^(2n/3) * (3³)^(-n/6) } / (75)^(-n/6)

using (a^m)^n = (a)^(m * n) in numerator now,

→ [5^{3 * (2n/3)} * 3^{3 * (-n/6)}] / (75)^(-n/6)

→ [5^(2n) * 3^(-n/2)] / (75)^(-n/6)

now, breaking the denominator part ,

→ [5^(2n) * 3^(-n/2)] / (25 * 3)^(-n/6)

→ [5^(2n) * 3^(-n/2)] / (5² * 3)^(-n/6)

using (a * b)^m = a^m * b^m in denominator, we get,

→ [5^(2n) * 3^(-n/2)] / [(5²)^(-n/6) * 3^(-n/6)]

→ [5^(2n) * 3^(-n/2)] / [5^{2 * (-n/6)} * 3^(-n/6)]

→ [5^(2n) * 3^(-n/2)] / [5^(-n/3) * 3^(-n/6)]

now, using a^m / a^n = a^(m - n) in both terms we get,

→ 5^{2n - (-n/3)} * 3^{(-n/2) - (-n/6)}

→ 5^{(6n + n)/3} * 3^{(-3n + n)/6}

→ 5^(7n/3) * 3^(-2n/6)

→ 5^(7n/3) * 3^(-n/3)

finally using a^(-m) = 1/a^m , we get,

→ [ 5^(7n/3) / 3^(n/3) ] (Ans.)

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