Question

\(\dfrac{1}{9}\) x \(27^n\) = 3ⁿ

\(\dfrac{8}{2^n}\) = 2

32ⁿ x 16^-n

3^-1 x 3ⁿ + 5 x 3^n-1 = 162

( n - \(\dfrac{2}{3}\) )³ = \(\dfrac{1}{27}\)

 

Answer 1

    \(\dfrac{1}{9}\).27ⁿ = 3ⁿ

   \(\dfrac{3^{3n}}{3^2}\) = 3ⁿ

   3^{3n -2} = 3ⁿ

     3n - 2 = n

    3n - n = 2

      2n = 2

        n = 2 : 2

        n = 1

Vậy n = 1

 

 

Answer 2

\(\dfrac{8}{2^n}\) = 2

\(\dfrac{2^3}{2^n}\) = 2

2^3-n = 2¹

3 - n = 1

     n = 3 -1 

     n = 2

Vậy n = 2

 

Answer 3

32ⁿ x 16^-n Thiếu vế phải

3^-1.3ⁿ + 5.3^n-1 = 162

3^n-1+ 5.3^n-1 = 162

3^n-1.(1 + 5) = 162

3^n-1.6  = 162

3^n-1 = 162 : 6

3^n-1 3= 27

3^n-1 = 3³

n - 1 = 3

n = 3 + 1

n = 4

Vậy n = 4

Answer 4

(n - \(\dfrac{2}{3}\))³ = \(\dfrac{1}{27}\)

(n - \(\dfrac{2}{3}\))³ = (\(\dfrac{1}{3}\))³

n - \(\dfrac{2}{3}\)   = \(\dfrac{1}{3}\)

 n = \(\dfrac{1}{3}\) + \(\dfrac{2}{3}\)

 n = 1

Vậy n = 1