Question

9n+2*(3-n/2)-2-27n/3m*2 3=1/729 then prove m-n=2

Answer 1

Question :-

If

$\large \dfrac{9^{n + 1} \times \left( 3^{ - \frac{n}{2}} \right)^{ - 2} - 27^n }{(3^m \times 2)^3 } = \dfrac{1}{729}$

then, prove that m - n = 2.

Solution :-

$\large \dfrac{9^{n + 1} \times \left( 3^{ - \frac{n}{2}} \right)^{ - 2} - 27^n }{(3^m \times 2)^3 } = \dfrac{1}{729}$

$\large \dfrac{(3^2)^{n + 1} \times \left( 3^{ - \frac{n}{2}} \right)^{ - 2} - (3^3)^n }{(3^m \times 2)^3 } = \dfrac{1}{3^6}$

[Because 9 can be written as 3² and 27 as 3³ and 729 as 3 to power of 6]

$\large \dfrac{3^{2(n + 1)} \times 3^{ - \frac{n}{2}( - 2)} - 3^{3(n)} }{(3^m \times 2)^3 } = \dfrac{1}{3^6}$

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

$\large \dfrac{3^{2n + 2} \times 3^{ - n( - 1)} - 3^{3n} }{(3^m \times 2)^3 } = \dfrac{1}{3^6}$

$\large \dfrac{3^{2n + 2} \times 3^n - 3^{3n} }{(3^m \times 2)^3 } = \dfrac{1}{3^6}$

It can be written as

$\large \dfrac{3^{2n} \times 3^2 \times 3^n - 3^{3n} }{(3^m \times 2)^3 } = \dfrac{1}{3^6}$

$\large \dfrac{3^{2n} \times 3^n \times 3^2 - 3^{3n} }{(3^m \times 2)^3 } = \dfrac{1}{3^6}$

$\large \dfrac{3^{2n + n} \times 3^2 - 3^{3n} }{(3^m \times 2)^3 } = \dfrac{1}{3^6}$

$\large \dfrac{3^{3n} \times 3^2 - 3^{3n} }{(3^m \times 2)^3 } = \dfrac{1}{3^6}$

$\large \dfrac{3^{3n} \times 3^2 - 3^{3n} }{(3^m)^3 \times {2}^{3} } = \dfrac{1}{3^6}$

$\bf \because {a}^{m} \times {b}^{m} = (ab)^{m}$

$\large \dfrac{3^{3n} \times 3^2 - 3^{3n} }{3^{3m} \times 8} = \dfrac{1}{3^6}$

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

Taking 3^{3n} common

$\large \dfrac{3^{3n}(3^2 -1) }{3^{3m} \times 8} = \dfrac{1}{3^6}$

$\large \dfrac{3^{3n}(9 -1) }{3^{3m} \times 8} = \dfrac{1}{3^6}$

$\large \dfrac{3^{3n}( \cancel 8) }{3^{3m} \times \cancel 8 } = \dfrac{1}{3^6}$

$\large \dfrac{3^{3n}}{3^{3m} } = \dfrac{1}{3^6}$

$\large \dfrac{3^{3n}}{3^{3m} } = {3}^{ - 6}$

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

$\large 3^{3n - 3m}= {3}^{ - 6}$

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

$\large 3n - 3m = - 6$

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

$\large 3(n - m) = - 6$

$\large n - m = - \dfrac{6}{3}$

$\large n - m = - 2$

$\large n - m + 2=0$

$\large n + 2= m$

$\large 2= m - n$

$\large \implies m - n = 2$

Hence proved

Answer 2

Refer to the above attachment :-)