Question

Solve this without spamming plz.​

Answer 1

$\boxed{\textbf{\large{to prove}}}$

If [(9^n x 3^2 x 3^n - ( 27 )^n )]/ [(3^(3m) x 2^3 ] = 3^( - 3 )

Then prove that ( m - n) = 1

$\boxed{\textbf{\large{Step-bystep explanation:}}}$

[(9^n x 3^2 x 3^n - ( 27 )^n )]/ [(3^(3m) x 2^3 ] = 3^( - 3 )

⟹ [ ( 9^n x 3^2 x 3^n - ( 27 )^n) ]

= ( 3^( - 3)x [ ( 3^( 3m) x 2^3 ) ]

⟹ [ 27^n x 9 - (27 )^n ]

= [ ( 3^(3m) x 8 ) / ( 3^(3) )]

⟹ [ ( 27^n ( 9 - 1 ) ]

= [ (( 27 )^(m) x 8 ) / 27 )]

⟹ 27^n x 8 = ((27^(m) x 8 )/ 27

⟹ 27^n = 27^m / 27

⟹ 27 = 27^m / 27^n

⟹ 27 = 27 ^m x 27^(- n)

⟹ 27^(1 ) = 27 ^( m - n)

Therefor, the bases of both sides are equal

( m - n) = 1

hence proved

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▶for such a types of problems ,we need to use certain identities given below.

◾( m / n) ^2 =( m )^2 / (n) ^2

◾ √( m / n) = √m / √n

◾( a ^(m )^(n)) = ( a^ ( m ✖ n))

◾( a ^( - m)) = 1 /( a ^(m) )

◾( a ^m ✖ a^n ) = ( a ^( m + n))

◾( 1 ) ^(z) = (1)

◾(a^(m)) /(a^(n)) = a^(m - n)

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