Question

Tolong bantuannya~
sekaligus cara kerjanya

Answer 1

Verified answer

3log (2^(x + 1) + 2^(2 - x)) = 2
3log (2^x . 2 + 2^2 . 2^-x) = 3log 3^2
2 . 2^x + 4 . 1/(2^x) = 3^2
Misal p = 2^x
2p + 4 . 1/p = 9 ======> kali p
2p^2 + 4 = 9p
2p^2 - 9p + 4 = 0
(2p - 1)(p - 4) = 0
p = 1/2 atau p = 4
2^x = 2^-1 atau 2^x = 2^2
x = -1 atau x = 2
Karena x1 < x2 maka x1 = -1 dan x2 = 2
x1 + x2 = -1 + 2 = 1

Answer 2

3log (2^(x + 1) + 2^(2 - x)) = 2
3log (2^x . 2 + 2^2 . 2^-x) =
2p + 4 . 1/p = 9
2p^2 + 4 = 9p
2p^2 - 9p + 4 = 0
(2p - 1)(p - 4) = 0

P1 = -1 atau p2 = 2
x1 + x2 = -1 + 2 = 1( C )