Question

Answer it properly please

Answer 1

Answer:

m - n = 1

Step-by-step explanation:

$\large \text{ \dfrac{9^n\times3^2\times\left((3)^{\dfrac{-n}{2}\right)^{-2}-27^{n}}}{3^{3m}\times2^3}=\dfrac{1}{27} }\\\\\\\large \text{ \implies \dfrac{3^{2n}\times3^2\times\left((3)^{\dfrac{-n}{2}\times{-2}\right)-3^{3n}}}{3^{3m}\times2^3} =\dfrac{1}{3^3} }\\\\\\\large \text{ \implies \dfrac{3^{2n}\times3^2\times3^{n}-3^{3n}}{3^{3m}\times2^3} =\dfrac{1}{3^3} }\\\\\\\large \text{Taking out 3^{3n} common}$

$\large \text{ \implies \dfrac{3^{3n}\left(3^2-1\right)}{3^{3m}\times2^3} =\dfrac{1}{3^3} }\\\\\\\large \text{ \implies \dfrac{3^{3n}\left(8\right)}{3^{3m}\times2^3} =\dfrac{1}{3^3} }\\\\\\\large \text{ \implies \dfrac{3^{3n}\left(2^3\right)}{3^{3m}\times2^3} =\dfrac{1}{3^3} }\\\\\\\large \text{\implies 2^3 get cancel out}\\\\\\\large \text{ \implies \dfrac{3^{3n}}{3^{3m}} =\dfrac{1}{3^3} }\\\\\\\large \text{ \implies \dfrac{1}{3^{3m}\times3^{3n}} =\dfrac{1}{3^3} }$

$\large \text{ \implies \dfrac{1}{3^{3m-3n}} =\dfrac{1}{3^3} }\\\\\\\large \text{Comparing both side we get}\\\\\\\large 3m-3n=3\\\\\\\large m-n=1$

Thus we get answer.

Answer 2

Question :-

If

$\large{ \sf \dfrac{9^n \times 3^2 \times \left(3^{ \frac{ - n}{2} } \right)^{ - 2} -27^{n} }{3^{3m} \times 2^3 } = \dfrac{1}{27} }$

then m - n

Answer :-

m - n = 1

Solution :-

$\large{ \sf \dfrac{9^n \times 3^2 \times \left(3^{ \frac{ - n}{2} } \right)^{ - 2} -27^{n} }{3^{3m} \times 2^3 } = \dfrac{1}{27} }$

$\large{ \sf \dfrac{(3^2)^n \times 3^2 \times \left(3^{ \frac{ - n}{2} } \right)^{ - 2} - (3^3)^n }{3^{3m} \times 2^3 } = \dfrac{1}{3^3} }$

[ Because 9 can be written as 3² , 27 as 3³]

$\large{ \sf \dfrac{3^{2(n)} \times 3^2 \times 3^{- n( - 1)}- 3^{3(n)}}{3^{3m} \times 2^3 } = \dfrac{1}{3^3} }$

$\bf \because (a^m)^n = a^{mn}$

$\large{ \sf \dfrac{3^{2n} \times 3^2 \times 3^{n}- 3^{3n}}{3^{3m} \times 2^3 } = \dfrac{1}{3^3} }$

$\large{ \sf \dfrac{3^{2n} \times 3^{n} \times 3^2 - 3^{3n}}{3^{3m} \times 2^3 } = \dfrac{1}{3^3} }$

$\large{ \sf \dfrac{3^{2n + n} \times 3^2 - 3^{3n}}{3^{3m} \times 2^3 } = \dfrac{1}{3^3} }$

$\bf \because (a^m)^n = a^{mn}$

$\large{ \sf \dfrac{3^{3n} \times 3^2 - 3^{3n}}{3^{3m} \times 2^3 } = \dfrac{1}{3^3} }$

Taking $\bf 3^n$ as common

$\large{ \sf \dfrac{3^{3n}(3^2 - 1)}{3^{3m} \times 8 } = \dfrac{1}{3^3} }$

$\large{ \sf \dfrac{3^{3n}(9 - 1)}{3^{3m} \times 8 } = \dfrac{1}{3^3} }$

$\large{ \sf \dfrac{3^{3n}( \cancel 8)}{3^{3m} \times \cancel 8 } = \dfrac{1}{3^3} }$

$\large{ \sf \dfrac{3^{3n}}{3^{3m} } = \dfrac{1}{3^3} }$

$\large{ \sf 3^{3n - 3m} = \dfrac{1}{3^3} }$

$\bf \because \dfrac{ a^{m} }{ {a}^{n} } = a^{m - n}$

$\large{ \sf 3^{3n - 3m} = 3^{ - 3}}$

$\bf \because \dfrac{1}{ {a}^{n} } = a^{- n}$

$\large{ \sf 3n - 3m = - 3}$

$\bf \because {a}^{m} = {a}^{n} \implies m = n$

$\large{ \sf n - m = - \dfrac{3}{3} }$

$\large{ \sf n - m = - 1}$

$\large{ \sf n = - 1 + m}$

$\large{ \sf 0 = - 1 + m - n}$

$\large{ \sf 1 = m - n}$

$\large{ \sf \implies m - n = 1}$

Laws of exponents used :-

$\tt \rightarrow (a^m)^n = a^{mn}$

$\tt \rightarrow \dfrac{ a^{m} }{ {a}^{n} } = a^{m - n}$

$\tt \rightarrow \dfrac{1}{ {a}^{n} } = a^{- n}$

$\tt \rightarrow {a}^{m} = {a}^{n} \implies m = n$