Question

find m-n with complete steps

Answer 1

{ 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

Answer 2

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