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Math Help - Solve the system

  1. #1
    Super Member dhiab's Avatar
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    Solve the system

    Solve the system :

    \left\{ \begin{array}{l}<br />
 x^y  = y^x  \\ <br />
 a^x  = b^y  \\ <br />
 \end{array} \right.
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  2. #2
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    I hope that you like logarithms. Hopefully there may be an easier solution than mine!

    From equation #2
    x=log_a(b^y)=ylog_ab

    Substituting for x in equation #1
    \therefore (ylog_ab)^y=y^{ylog_ab}

    \therefore y^y(log_ab)^y=(y^y)^{log_ab}

    \therefore (log_ab)^y=(y^y)^{(log_ab-1)}

    taking log of both sides to base log_ab

    \therefore log_{log_ab}((log_ab)^y)=y=log_{log_ab}(y^y)^{(log  _ab-1)}=y(log_ab-1)log_{log_ab}(y)

    dividing through by y

    \therefore (log_ab-1)log_{log_ab}(y)=1

    \therefore log_{log_ab}(y)=\frac 1{log_ab-1}

    \therefore y=log_ab to the power of  (\frac 1{log_ab-1})

    Example; say a = e =2.718 and b = 17 then y=1.764875.

    x=log_a(b^y)=5.000267 from equation #2

    finally

    x^y=y^x=17.1252
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  3. #3
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    Quote Originally Posted by dhiab View Post
    Solve the system :
    \left\{ \begin{array}{l}<br />
x^y = y^x \\ <br />
a^x = b^y \\ <br />
\end{array} \right.
    Why can't x = y = 2 and a = b = any number n ?
    2^2 = 2^2
    n^2 = n^2
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  4. #4
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    It is usual for letters from the first half of the alphabet to be considered as constants and letters from the last half of the alphabet to be variables. Therefore, if the question had been written more formally it might have said something like:

    Solve the following system of equations for x and y, given the constants a and b.
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  5. #5
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    Quote Originally Posted by Kiwi_Dave View Post
    It is usual.....
    That to me means no more than "it is usual for men to wear pants
    and wowen to wear dresses".
    "It is usual" is quite dangerous in mathematics...
    In programming, using a to m as variables is not "syntax error".

    Ain't arguing, mate, just my opinion.
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