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    • Thread Starter

    How do i solve these equations... Can i have a step by step answer as i missed it in class

    1. 2(3^2x) - 7(3^x) + 6 =0
    2. log base 4x - 3log base x4= 2

    1.Note that 3^{2x} = (3^x)^2, so your equation reduces to 2u^2 - 7u + 6 = 0 where u = 3^x.

    2. Do you know the base changing formula? (If not, search for it; I've forgotten it.)

    (Original post by ghostkillah)
    log base 4x - 3log base x4= 2
    log_ab \equiv \frac {1}{log_ba}
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