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9^x+2 - 6* 3^x+1= 0 find x

Answer earlier I got confused

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  1. I believe it was Steve who had answered you earlier.
    He started by letting u = 3^x

    look at the first term 9^(x+2) <--- I will assume those necessary brackets, as did Steve.
    9^(x+2) = (3^2)^(x+2)
    = 3^(2x + 4)
    = 3^2x * 3^4
    = 81 * (3^x)^2 , also 6*3^(x+1) = 6*3^x * 3^1 = 18*3^x

    so, 9^(x+2) - 6* 3^(x+1) = 0 becomes

    81 * (3^x)^2 - 18*3^x = 0
    81 u^2 - 18u = 0
    9u(9u - 2) = 0
    9u = 0 ---> u = 0
    or
    9u = 2 ---> u = 2/9

    so 3^x = 0 , which has no solution

    OR

    3^x = 2/9
    xlog3 = log2 - log9
    x = (log2 - log9)/log3 = appr -1.369

    This answer can be verified in the original equation

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