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Math Help - Help with simple equation needed.

  1. #1
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    Help with simple equation needed.

    Hi,

    I cannot solve the following equation to x.

    27^{\sqrt{2-3x}} = 243^{\sqrt{3x-2}}

    I tried it like this:

    \sqrt{2-3x}lg(27) = \sqrt{3x-2}lg(243)

    after some squaring and multiplying I got the following:

    2-3x = \frac{3x(lg(243))^2} {(lg(27))^2} - \frac{2(lg(243))^2)}{(lg(27))^2}

    Then I didn't know how to continue. Wolfram Alpha says that x is 2/3: http://www.wolframalpha.com/input/?i=2-3x+%3D+%28lg%28243%29%2F%28lg%2827%29%29%29^2*%283 x-2%29

    My textbook, from where I have this equation, also says x = 2/3. But I don't know how to get there. I hope that someone can post the last steps to solve to x.
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  2. #2
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    Quote Originally Posted by Antamara View Post
    Hi,

    I cannot solve the following equation to x.

    27^{\sqrt{2-3x}} = 243^{\sqrt{3x-2}}

    I tried it like this:

    \sqrt{2-3x}lg(27) = \sqrt{3x-2}lg(243)

    after some squaring and multiplying I got the following:

    2-3x = \frac{3x(lg(243))^2} {(lg(27))^2} - \frac{2(lg(243))^2)}{(lg(27))^2}

    Then I didn't know how to continue. Wolfram Alpha says that x is 2/3: http://www.wolframalpha.com/input/?i=2-3x+%3D+%28lg%28243%29%2F%28lg%2827%29%29%29^2*%283 x-2%29

    My textbook, from where I have this equation, also says x = 2/3. But I don't know how to get there. I hope that someone can post the last steps to solve to x.
    Why didn't you try

    3^{3\sqrt{2-3x}}=3^{5\sqrt{3x-2}} ?

    Then equate the indices and square both of them
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  3. #3
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    Do you "see" (from Archie's reply) that 3SQRT(2 - 3x) = 5SQRT(3x - 2) ?
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  4. #4
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    Thank you very much for the fast answers. I tried it with Archie Meades' recommendation and it works. I also tried to continue from here:

    2-3x = \frac{3x(lg(243))^2} {(lg(27))^2} - \frac{2(lg(243))^2)}{(lg(27))^2}

    I transformed the above equation to this:

    x = (\frac{-2(lg243)^2}{(lg27)^2}-2) : (\frac{-3(lg(243))^2} {(lg27)^2}-3)

    Which gives x = 2/3.
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  5. #5
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    Notice that, as Archie Meade pointed out, 27= 3^3 while 243= 3^5. That means that \frac{log(243)}{log(27)}= \frac{log(3^5)}{log(3^3)} = \frac{5 log(3)}{3 log(3)}= \frac{5}{3}.
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