Question

9^n*3^2*3n-(27)^n/
(3^15*2^3=1/27​

Answer 1

Given that, ( 9^n x 3^2 x (3^-n/2)^-2 - 27^n) /( 3^3m x 2^3) = 1/27

⇒ (3^2n x 3^2 x { 3^-(-2n/2) } - 3^3n )/ 3^3m x 2^3 = 1/(3^3)

⇒ (3^2n x 3^2 x {3^n} - 3^3n )/ 3^3m x 2^3 = 3^(-3)

⇒ ( 3^3n x 3^2 - 3^3n )/ 3^3m x 2^3 = 3^(-3)

⇒ 3^3n[ 3^2 - 1] / 3^3m x 2^3 = 3^(-3)

⇒ 3^3n[ 9 - 1 ] /3^3m x 8 = 3^(-3)

⇒ 3^3n x 8 /3^3m x 8 = 3^(-3)

⇒ 3^(3n-3m) = 3^(-3)

comparing powers on both side, we get

3n - 3m = -3

⇒ 3(n-m) = -3

⇒ (n-m) = -1

or m-n = 1

Hence proved...Show more

0

Answer 2

$\Large{\underline{\underline{\red{\mathfrak{Given:-}}}}}$

$\rm{ \frac{ {9}^{n} \times {3}^{2} \times {3}^{n} - {27}^{n} }{ {3}^{3m} \times {2}^{3} } = \frac{1}{27} }$

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$\Large{\underline{\underline{\orange{\mathfrak{To \: find:-}}}}}$

$\rm{m-n=1}$

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$\Large{\underline{\underline{\pink{\mathfrak{Solution:-}}}}}$

⟾ $\rm{ \frac{ {9}^{n} \times {3}^{2} \times {3}^{n} - {27}^{n} }{ {3}^{3m} \times {2}^{3} } = \frac{1}{27} }$

⟾ $\rm{ \frac{ ( { ({3}^{2} )}^{n} \times {3}^{2} \times {3}^{n} \times - { ({3}^{3}) }^{n} }{ {3}^{3m} \times {2}^{3} } = \frac{1}{27} }$

⟾ $\rm{ \frac{ {3}^{2n} \times {3}^{2} \times {3}^{n} - {3}^{3n} }{ {3}^{3m} \times {2}^{3} } = \frac{1}{27} }$

⟾ $\rm{ \frac{ {3}^{2n + 2 + n} - {3}^{3n} }{ {3}^{3m} \times {2}^{3} } = \frac{1}{ {3}^{3} } }$

⟾ $\rm{ \frac{ {3}^{3n + 2} - {3}^{n} }{ {3}^{m} \times {2}^{3} } = \frac{1}{ {3}^{3} } }$

⟾ $\rm{ \frac{ {3}^{3n } \times {3}^{2} - {3}^{n} }{ {3}^{m} \times {2}^{3} } = \frac{1}{ {3}^{3} } }$

⟾ $\rm{ \frac{ {3}^{3n} ( {3}^{2} - 1) }{ { 3}^{3m} \times 8 } = \frac{1}{ {3}^{3} } }$

⟾ $\rm{ \frac{ {3}^{3n} \times 8 }{ { 3}^{3m} \times 8 } = \frac{1}{ {3}^{3} } }$

⟾ $\rm{ {3}^{3n - 3m} = {3}^{ - 3} }$

⟾ $\rm{3n-3m=-3}$

⟾ $\rm{n-m=-1}$

⟾ $\rm{m-n=1}$

${\bold{\underline{\underline{\green{\mathfrak{Hence\: Verified!}}}}}}$

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$\Large{\underline{\underline{\blue{\mathfrak{Thanks}}}}}$