Matematika

Pertanyaan

tolong bantuannya~
sekaligus cara kerjanya
tolong bantuannya~ sekaligus cara kerjanya

2 Jawaban

  • 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
  • 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 )