Answer:

SOLUTION

:-

\begin{gathered}\bf = > \frac{\red{ {9}^{n} \times {3}^{2} \times ( { {3}^{ \frac{ - n}{\cancel2} } })^{ - \cancel 2} - {27}^{n}}}{\red{ {3}^{3m} \times {2}^{3} }} = \frac{\red{1}}{\red{27}} \\ \\ \bf= > \frac{\red{( {3})^{2n} \times {(3)}^{2} \times {(3)}^{n} - ( {27})^{n} }}{\red{ ({3})^{3m} \times ({2})^{3} } } = \frac{\red{1}}{\red{27}} \end{gathered}

=>

3

3m

×2

3

9

n

×3

2

×(3

2

−n

)

−

2

−27

n

=

27

1

=>

(3)

3m

×(2)

3

(3)

2n

×(3)

2

×(3)

n

−(27)

n

=

27

1

since, base are common ; i.e, (3) is common and they are multiplying, so there exponents will add such that :

\begin{gathered}\bf = > \frac{\red{( {3})^{2n + 2 + n} - {(3)}^{3N}}}{\red{({3})^{m} \times ({2})^{3} }} = \frac{\red{1}}{\red{27}} \\ \\ \bf = > \frac{\red{ {(3)}^{3n + 2} - {(3)}^{3n} }}{ \red{{(3)}^{3m} \times {(2)}^{3} } } = \frac{\red{1}}{\red{27} }\end{gathered}

=>

(3)

m

×(2)

3

(3)

2n+2+n

−(3)

3n

=

27

1

=>

(3)

3m

×(2)

3

(3)

3n+2

−(3)

3n

=

27

1

now , take {(3)}^{3n}(3)

3n

common from numerator!

\begin{gathered}\bf = > \frac{\red{ {(3)}^{3n}( {3}^{2} - 1) }}{ \red{{(3)}^{3m} \times {(2)}^{3} }} = \frac{\red{1}}{\red{27}} \\ \\ \bf = > \frac{ \red{{(3)}^{3n}(9 - 1) }}{\red{ {(3)}^{3m} \times 8} } = \frac{\red{1}}{\red{27}} \\ \\ \bf = > \red{\frac{ {(3)}^{3n} \times \cancel8 }{ {(3)}^{3m} \times \cancel8}} = \red{\frac{1}{27} }\end{gathered}

=>

(3)

3m

×(2)

3

(3)

3n

(3

2

−1)

=

27

1

=>

(3)

3m

×8

(3)

3n

(9−1)

=

27

1

=>

(3)

3m

×

8

(3)

3n

×

8

=

27

1

then, again (3) is common, but it is dividing, so exponents will subtract :

\begin{gathered}\bf = > \red{{(\cancel3)}^{3n - 3m}} = \red{( {\cancel3)}^{ - 3}} \\ \\ \bf= > \red{ 3n - 3m = - 3}\end{gathered}

=>(

3

)

3n−3m

=(

3

)

−3

=>3n−3m=−3

take 3 as common from LHS ,

\bf = > \red{3(n - m) = - 3}=>3(n−m)=−3

take 3 to RHS from LHS , as it's multiplying, so taking other side, it will divide ;

\begin{gathered}\bf = > \red{n - m = - \frac{3}{3}} \\ \\ \bf = > \red{ n - m = - 1} \\ \\ \bf = > \red{\cancel - (m - n) = \cancel- 1} \\ \\ \bf = > \blue{\boxed{\huge{m - n = 1}}}\end{gathered}

=>n−m=−

3

3

=>n−m=−1

=>

−

(m−n)=

−

1

=>

m−n=1

★ HENCE PROVED !! ★