Algebra logs please help

Log3(x)=log9(6x)

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  1. since 9 = 3^2,
    log3(n) = 2log9(n)
    That is, the power of 3 needed for a number is twice as big as the power of 9.

    So, we have

    2log9(x) = log9(6x)
    log9(x^2) = log9(6x)
    x^2 = 6x
    x = 0 or 6
    0 not allowed, so x=6

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