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Math Help - Indices (simple but i am have problems)

  1. #1
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    Indices (simple but i am getting problems)

    9^x+1 - 13(3^x) + 4 =0

    Use y = 3^x

    OK so i start and reached to this point
    3^2(x+1) - 13(3^x) + 4 = 0


    3^2x+2 - 13(3^x) + 4 = 0

    I need some help to proceed, thanks in advance
    Last edited by kingmaths; January 8th 2012 at 09:32 AM.
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  2. #2
    Super Member Bacterius's Avatar
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    Re: Indices (simple but i am have problems)

    3^{2x + 2} - 13 \times 3^x + 4 = 0

    Expand:

    3^{2x} \times 3^2 - 13 \times 3^x + 4 = 0

    Which becomes (from the rules of indices - look at the first term):

    \left (3^{x} \right )^2 \times 3^2 - 13 \times 3^x + 4 = 0

    Now use the hint and substitute y= 3^x

    y^2 \times 3^2 - 13y + 4 = 0

    So:

    9 y^2 - 13y + 4 = 0

    Can you solve for y and put the solution back in terms of x now?
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