if 9^n*3^2*3^n-27^n/3^3m*2^3=1/27 prove that m-n=1
Answers
Answer:
SOLUTION
:-
\begin{lgathered}\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{lgathered}
=>
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{lgathered}\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{lgathered}
=>
(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{lgathered}\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{lgathered}
=>
(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{lgathered}\bf = > \red{{(\cancel3)}^{3n - 3m}} = \red{( {\cancel3)}^{ - 3}} \\ \\ \bf= >\red{ 3n - 3m = - 3}\end{lgathered}
=>(
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{lgathered}\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{lgathered}
=>n−m=−
3
3
=>n−m=−1
=>
−
(m−n)=
−
1
=>
m−n=1
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