(-2^x + 99) [-2^(x-1)] = 48, how do you solve this?

If someone can answer in step by step that would be helpful

Update:

@mike G how did you get to that first step

2 Answers

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  • Pranil
    Lv 7
    8 years ago
    Favorite Answer

    (-2^x + 99) [-2^(x-1)] = 48

    (-2^x + 99) [-2^x/2^1)] = 48

    (-2^x + 99) [-2^x] = 96

    4^x – 99(2^x) = 96

    2^2^x = 99(2^x) + 96 --------- Let 2^x = a

    a^2 – 99a + 96 = 0

    a = [99 ± √99² – 384]/2

    a = 98.01 or 0.98

    2^x = 98.01 or 2^x = 0.98

    x log 2 = log 98.01

    x = 1.9913/0.3010 = 6.62

    or

    x = – 0.02915

    ----

  • Mike G
    Lv 7
    8 years ago

    2^(2x-1)-99*2^(x-1)-48 = 0

    2^2x-99*2^x - 96 = 0

    2^x = [99±sqrt(10185)]/2

    2^x = 99.96

    xlog2 = log99.96

    x = 6.6433

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