Question

If 9^n*3^2*3^n-27^n/3^3m*2^3 = 1/27, prove that m-n=1.

Answer 1

here is the answer of your given question

Answer 2

$\mathcal{\underline{SOLUTION}}$ :-

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

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

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

now , take ${(3)}^{3n}$ common from numerator!

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

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

$\bf = > \red{{(\cancel3)}^{3n - 3m}} = \red{( {\cancel3)}^{ - 3}} \\ \\ \bf= >\red{ 3n - 3m = - 3}$

take 3 as common from LHS ,

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

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

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

★ HENCE PROVED !! ★