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Math Help - Solve for x...

  1. #1
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    Solve for x...

    Hi,

    I'm not getting this. I'm wrong - as usual. May I have some help please?

    If 3(27)^x = 9^{2x\,+\,1}, what is the value of x?

    81^x = 9^{2x\,+\,1}

    x \log 81 = (2x + 1) \log 9

    x \log 81 = 2x \log 9 + \log 9

    x \log 81 - 2x \log 9 = \log 9

    x(\log 81 - 2 \log 9) = \log 9

    x = \frac{\log 9}{\log 81 - 2 \log 9}

    x = Undefined
    Last edited by Hellbent; November 23rd 2010 at 11:09 PM.
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  2. #2
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    I can't read the power on the RHS, but remember that \displaystyle 27 = 3^3 and \displaystyle 9=3^2.

    Convert everything to powers of \displaystyle 3 and then you can equate the powers.
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  3. #3
    Senior Member Educated's Avatar
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    I don't think are any real answers for x.

    Edit: Sorry, I edited it a minute after because I didn't think anyone would reply so soon.
    The only solution I can find was -1. But by calculations I cannot seem to find any other answers.
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  4. #4
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    \displaystyle 3\cdot 27^x = 9^{2x+1}

    \displaystyle 3(3^3)^x = (3^2)^{2x+1}

    \displaystyle 3\cdot 3^{3x} = 3^{2(2x+1)}

    \displaystyle 3^{3x+1} = 3^{4x+2}

    \displaystyle 3x+1=4x+2

    \displaystyle 1 = x+2

    \displaystyle x= -1.
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  5. #5
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    Quote Originally Posted by Prove It View Post
    I can't read the power on the RHS,...
    Sorry, I fixed it.

    Quote Originally Posted by Prove It View Post
    but remember that \displaystyle 27 = 3^3 and \displaystyle 9=3^2.

    Convert everything to powers of \displaystyle 3 and then you can equate the powers.
    3(27)^x = 9^{2x\,+\,1}

    3(3^3)^x = (3^x)^{2x\,+\,1}

    3(3^{3x}) = (3)^{4x\,+\,2}

    3 (3x \log 3) = 4x + 2(\log 3)

    9x \log 9 = 4x \log 3 + 2\log 3

    9x \log 9 - 4x \log 3 = 2 \log 3

    x(9 \log 9 - 4 \log 3) = 2 \log 3

    x = \frac{2 \log 3}{9 \log 9 - 4 \log 3}

    x \approx -1.797
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  6. #6
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    Quote Originally Posted by Educated View Post
    The only solution I can find was -1. But by calculations I cannot seem to find any other answers.
    Yes, the answer is -1.

    Thanks.
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  7. #7
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    Quote Originally Posted by Hellbent View Post
    Sorry, I fixed it.



    3(27)^x = 9^{2x\,+\,1}

    3(3^3)^x = (3^x)^{2x\,+\,1}

    3(3^{3x}) = (3)^{4x\,+\,2}

    3 (3x \log 3) = 4x + 2(\log 3)

    9x \log 9 = 4x \log 3 + 2\log 3

    9x \log 9 - 4x \log 3 = 2 \log 3

    x(9 \log 9 - 4 \log 3) = 2 \log 3

    x = \frac{2 \log 3}{9 \log 9 - 4 \log 3}

    x \approx -1.797
    If you want to use logarithms, then at this step

    \displaystyle 3(3^{3x}) = 3^{4x+2}

    you need to take the logarithm of both sides

    \displaystyle \log{[3(3^{3x})]} = \log{(3^{4x+2})}

    \displaystyle \log{3} + \log{(3^{3x})} = \log{(3^{4x+2})}

    \displaystyle \log{3} + 3x\log{3} = (4x+2)\log{3}

    \displaystyle (3x+1)\log{3} = (4x+2)\log{3}

    \displaystyle 3x+1=4x+2


    Go from there...
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  8. #8
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    Quote Originally Posted by Prove It View Post
    \displaystyle 3x+1=4x+2

    Go from there...
    3x - 4x = 2 - 1

    -x = 1

    x =  -1
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