Question

Tìm x biết :

a) \(2^{-1}.2^x+4.2^x=9.2^5\)

b) \(\left(7x+2\right)^{-1}=\frac{1}{9}\)

Answer 1

a) 2^-1.2^x + 4.2^x = 9.2⁵

=> 2^x-1 + 4.2.2^x-1 = 9.2⁵

=> 2^x-1 + 8.2^x-1 = 9.2⁵

=> 2^x-1.(1 + 8) = 9.2⁵

=> 2^x-1.9 = 9.2⁵

=> x - 1 = 5

=> x = 5 + 1 = 6

Vậy x = 6

b) (7x + 2)^-1 = \(\frac{1}{9}\)

=> 7x + 2 = 9

=> 7x = 9 - 2

=> 7x = 7

=> x = 7 : 7 = 1

Vậy x = 1

Answer 2

a)     2^(-1) . 2^x + 4 . 2^x = 9 . 2^5

=>    1/2 . 2^x + 4 . 2^x     = 9 . 32

=>    2^x . (1/2 + 4)           = 288

=>   2^x . 9/2 = 288 

=> 2^x = 288 : 9/2

=> 2^x = 64

      2^x = 2^6

=> x = 6

b) (7x + 2)^(-1) = 1/9

=> 1 phần 7x + 2  = 1/9

=> 7x + 2 = 9

=> 7x = 9 - 2 

=> 7x = 7

=> x = 7 : 7

=> x = 1