Math

Solve for the inequality:

1/3^x (1/3^x - 2) < 15

I know the answer is x > - (log 5/log 3), can someone explain me how?

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asked by Jam
  1. suppose we just look at
    1/3^x (1/3^x - 2) = 15
    3^-x(3^-x - 2) = 15
    let 3^-x = a for easier typing
    then we have
    a(a-2) = 15
    a^2 - 2a - 15 = 0
    (a-5)(a+3) = 0
    a = 5 or a = -3

    3^-x = 5
    log 3^-x = log5
    -x log3 = log5
    -x = log5/log3
    x = -log5/log3

    the other part of 3^-x = -3 has no solution

    but we wanted
    1/3^x (1/3^x - 2) < 15
    so x < -log5/log3

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    posted by Reiny

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