(3^n×9^n+1)÷(3^n-1×27^n-1)=81
Answers
Answer:
(3^n×9^n+1)÷(3^n-1×27^n-1)=81
(3^3n+2)÷(3^4n-4)=81
3n+2-4n+4=4
6-n=4
-n=-2
n=2
The value of n in the given equation is 2.
The given equation is :
( 3ⁿ × 9ⁿ ⁺ ¹ ) ÷ ( 3ⁿ ⁻ ¹ × 27ⁿ ⁻ ¹ ) = 81
=> [ 3ⁿ × (3²)ⁿ ⁺ ¹ ] ÷ [ 3ⁿ ⁻ ¹ × (3³)ⁿ ⁻ ¹ ] = 81
[ Writing 9 as 3², and 27 as 3³ in the equation ]
=> [ 3ⁿ × (3²ⁿ ⁺ ²) ] ÷ [ 3ⁿ ⁻ ¹ × (3³ⁿ ⁻ ³ ] = 81
[ By applying the property of (aˣ)ⁿ = aˣⁿ ]
=> [ 3ⁿ ⁺ ⁽²ⁿ ⁺ ²⁾ ] ÷ [ 3⁽ⁿ ⁻ ¹⁾ + ⁽³ⁿ - ³⁾ ] = 81
[ By applying the property of aˣ × aⁿ = aˣ ⁺ ⁿ ]
=> [ 3ⁿ ⁺ ²ⁿ ⁺ ² ] ÷ [ 3ⁿ ⁻ ¹ ⁺ ³ⁿ ⁻ ³ ] = 81
=> [ 3³ⁿ ⁺ ² ] ÷ [ 3⁴ⁿ ⁻ ⁴ ] = 81
=> [ 3⁽³ⁿ ⁺ ²⁾ ⁻ ⁽⁴ⁿ ⁻ ⁴⁾ ] = 81
[ By applying the property of aˣ ÷ aⁿ = aˣ - ⁿ ]
Now, 81 can be written as 3⁴.
=> 3⁽³ⁿ ⁺ ² ⁻ ⁴ⁿ ⁺ ⁴⁾ = 3⁴
=> 3n + 2 - 4n + 4 = 4
[ By applying the property of equal bases having equal powers on both sides, aˣ = aⁿ = > x = n ]
=> 3n - 4n + 2 + 4 = 4
=> - n + 6 = 4
=> - n = 4 - 6
=> - n = - 2
=> n = 2