find m-n with complete steps
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{ 9ⁿ (3²) (3^n/2)^-2 -27ⁿ}/{3^3m(2)³ } = 1/27
{ 3²ⁿ(3)² (3^n/2×-2) -3³ⁿ}/3^3m(2)³ =1/27
{3^(2n+2-n)- 3^3n}/8(3^3m} = 1/27
{3^(n +2) -3^3n} /3^3m = 8/27
here we see , can't solve by method ,
so, we should be solve by assumption method .
in LHS denominator = 3^3m
and RHS denominator = 27
for LHS = RHS
we let
3^3m = 27
3^3m = 3³
3m = 3
m = 1
similarly ,
in LHS numerator = {3^(n +2) -3^3n }
in RHS numerator = 8
for LHS = RHS
3^(n +2) -3^3n = 8
we let , n = 0
3^2 -3^0 = 9 -1 = 8
hence ,
m = 1 and n = 0
so, m - n = 1- 0 = 1
{ 3²ⁿ(3)² (3^n/2×-2) -3³ⁿ}/3^3m(2)³ =1/27
{3^(2n+2-n)- 3^3n}/8(3^3m} = 1/27
{3^(n +2) -3^3n} /3^3m = 8/27
here we see , can't solve by method ,
so, we should be solve by assumption method .
in LHS denominator = 3^3m
and RHS denominator = 27
for LHS = RHS
we let
3^3m = 27
3^3m = 3³
3m = 3
m = 1
similarly ,
in LHS numerator = {3^(n +2) -3^3n }
in RHS numerator = 8
for LHS = RHS
3^(n +2) -3^3n = 8
we let , n = 0
3^2 -3^0 = 9 -1 = 8
hence ,
m = 1 and n = 0
so, m - n = 1- 0 = 1
abhi178:
i solved this way only for finding integer value , by the way it can solve by logrithmn method but answer will be complicated
Answered by
1
Answer:
{ 9ⁿ (3²) (3^n/2)^-2 -27ⁿ}/{3^3m(2)³ } = 1/27
{ 3²ⁿ(3)² (3^n/2×-2) -3³ⁿ}/3^3m(2)³ =1/27
{3^(2n+2-n)- 3^3n}/8(3^3m} = 1/27
{3^(n +2) -3^3n} /3^3m = 8/27
here we see , can't solve by method ,
so, we should be solve by assumption method .
in LHS denominator = 3^3m
and RHS denominator = 27
for LHS = RHS
we let
3^3m = 27
3^3m = 3³
3m = 3
m = 1
similarly ,
in LHS numerator = {3^(n +2) -3^3n }
in RHS numerator = 8
for LHS = RHS
3^(n +2) -3^3n = 8
we let , n = 0
3^2 -3^0 = 9 -1 = 8
hence ,
m = 1 and n = 0
so, m - n = 1- 0 = 1
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