equation

9^x(-2)+3^3(x)=81 i need help

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  1. If you mean
    (9^x)^(-2) + (3^3)^x = 81
    then you have

    3^(-4x) + 3^(3x) = 81
    if u = 3^x, then you have

    1/u^4 + u^3 = 81

    I'm sorry, but I don't see any algebraic solution. you sure there's no typo? Wolframalpha gets this:

    http://www.wolframalpha.com/input/?i=3^%28-4x%29+%2B+3^%283x%29+%3D+81

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  2. steve can logarithm work 4 dat

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  3. I did similar what Steve did, and agree with him

    Steve asked you to improve your typing of the equation, but this does not change things.

    I will assume you meant:
    9^(-2x) + 3^(3x) = 81
    which becomes
    3^(-4x) + 3^(3x) = 81
    (3^-x)^4 + (3^-x)^-3 = 81
    let 3^-x = t
    so we have:
    t^4+ t^-3 = 81
    times t^3 for everybody
    t^7 + 1 = 81t^3

    tough to solve, unless we use something like Wolfram
    http://www.wolframalpha.com/input/?i=t%5E7+%2B+1+%3D+81t%5E3

    notice t = appr 3 or t = appr -3 are solutions to our last equation,
    so 3^-x = 3 = 3^1
    x = -1
    similarly x = appr +1

    somehow I don't think that is what you meant.

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  4. yes reiny daz the question i jus decided to include bracket in it,reiny u got me confuse how come 3^x become 3^-x,but what if i use logarithm can it work?

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  5. You must be referring to the 2nd term of
    (3^-x)^4 + (3^-x)^-3 = 81

    notice in the previous line it was 3^(3x)
    which is (3^x)^3 or (3^(-x))^-3

    Steve substituted 3^x, I substituted 3^-x,
    our equations reflect that

    No, logs will not work here, since you have to take the log of a sum, and we have no formulas to take care of log (A + B)

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