Math

logx(9x^2)[log3(x)]^2=4

i got the x=1/9 and x=3 is that correct?

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  1. Hard to say. Your syntax is murky. How did you arrive at your answer?
    You appear to have log(base x) of 9x^2 or log_10(x) * 9x^2
    and log(base 3) of x
    And what happens when you multiply logs?
    When you divide logs, you change base...

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  2. If the question is
    logx(9x^2)[log3 (x)]^2=4
    then x = 3 , 1/9 satisfies the equation.

    As oobleck asked, how did you get that, other than by inspection.

    I went as far as
    (2 log3/logx + 2)(logx/log3)^2 = 4

    when x = 3, the first bracket becomes (4) and the second bracket becomes 1, so it works
    if x = 1/9, the first bracket becomes (1) and the 2nd becomes (-2)^2, so it works

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    Reiny
  3. Nice work, Reiny. Using u = log3/logx, we could have proceeded with
    (2u+2)/u^2 = 4
    u+1 = 2u^2
    2u^2-u-1 = 0
    (2u+1)(u-1) = 0
    u = -1/2 or 1
    So, that means that log3/logx = -1/2 or 1
    logx/log3 = -2 or 1
    logkx = -2 or 1
    x = 3^-2 or 3^1
    x = 1/9 or 3

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