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Math Help - Solve The Equation !

  1. #1
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    Solve The Equation !

    Hii !

    solve in \mathbb{R} the equation :

    \frac{8^x+27^x}{12^x+18^x} = \frac{7}{6}
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  2. #2
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    Observation

    Note that 8 is 2 cubed, 27 is 3 cubed, 12 is 2 times 6 and 18 is 3 x 6. Also I would cross multiply and note what I just said in the prior sentence to try to work out a solution.
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  3. #3
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    Quote Originally Posted by wonderboy1953 View Post
    Note that 8 is 2 cubed, 27 is 3 cubed, 12 is 2 times 6 and 18 is 3 x 6. Also I would cross multiply and note what I just said in the prior sentence to try to work out a solution.


    Its True My Solution ???
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  4. #4
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    Quote Originally Posted by Perelman View Post


    Its True My Solution ???
    I agree with this up to the line (2^x+3^x)^2 = 25\times6^{x-1} and (2^x-3^x)^2 = 6^{x-1}. This clearly implies the following line, but is not equivalent to it. In fact, (2^x+3^x)^2 - (2^x-3^x)^2 = 24\times6^{x-1} appears to be a tautology, valid for all x. I certainly don't understand how it implies that 2\times3^x = 24\times6^{x-1}.

    Starting from (2^x+3^x)^2 = 25\times6^{x-1} and (2^x-3^x)^2 = 6^{x-1} I would take the positive square root on both sides of those equations. The first one gives 2^x+3^x = 5\times 6^{(x-1)/2}. For the second equation we need to be more careful. If x>0 then 3^x>2^x and we get 3^x-2^x = 6^{(x-1)/2}. Thus the two equations together give 2^x+3^x = 5(3^x-2^x), from which 3^{x-1}=2^{x-1} and therefore x=1.

    But if x<0 then the positive square root of (2^x-3^x)^2 is 2^x-3^x, and a similar procedure to the previous paragraph gives 2^{x+1} = 3^{x+1}, so that x=1.

    Thus there are two solutions, x = 1 and x = 1.
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