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Math Help - [SOLVED] Eponential Equations

  1. #1
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    [SOLVED] Eponential Equations

    I need to solve for x in the equation:
    9^(2x+1)=81(27^2)
    I've tried various things, getting every answer but the correct one, which is 2. I have no idea how x=2 though.
    Here's one of my (futile) tries:
    9^(2x+1)=9^2(27^x)
    x+1=27^x
    1 = (27^xx)
    Guidance through the proper steps would be appreciated, thank you!
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  2. #2
    Junior Member
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    Quote Originally Posted by ria3 View Post
    I need to solve for x in the equation:
    9^(2x+1)=81(27^2)
    I've tried various things, getting every answer but the correct one, which is 2. I have no idea how x=2 though.
    Here's one of my (futile) tries:
    9^(2x+1)=9^2(27^x)
    x+1=27^x
    1 = (27^xx)
    Guidance through the proper steps would be appreciated, thank you!
    So:
    9^(2x+1)=81x 27^2

    Put everything as powers of 3:
    (3^2)^(2x+1)=(3^4)(3^3)^2
    = 3^(4x+2)=(3^4)(3^6)
    = 3^(4x+2)=3^10
    therefore: 4x+2=10, 4x=8, x=2

    Sorry for the strange powers, I'm still trying to get the hang of this math code stuff. =)
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  3. #3
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    thank you!
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