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Math Help - Exponential simultaneous equations

  1. #1
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    Exponential simultaneous equations

    A little help with these would be of help!

    i) x^y= y^x
    x^2=y^3

    Solve for x and y.

    ii)a^x=x^y
    a^y=x^x

    Same here.
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  2. #2
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    Hello Arka

    Welcome to Math Help Forum!
    Quote Originally Posted by Arka View Post
    A little help with these would be of help!

    i) x^y= y^x
    x^2=y^3

    Solve for x and y.

    ii)a^x=x^y
    a^y=x^x

    Same here.
    By inspection, we can see that one solution is \displaystyle x = y = 1. But if \displaystyle y \ne 1, then we can take logs, knowing that \displaystyle \log y \ne 0:
    \displaystyle \left\{\begin {array}{l} x^y = y^x\\<br />
x^2=y^3\end{array}
    \displaystyle \Rightarrow \left\{\begin{array}{l} y\log x = x\log y\\<br />
2\log x = 3 \log y \end{array}

    \displaystyle \Rightarrow  \dfrac{\log x}{\log y}=\dfrac xy=\dfrac32 , provided \displaystyle \log y \ne 0.

    \displaystyle \Rightarrow y = \tfrac23x

    \displaystyle \Rightarrow y^3 = \tfrac{8}{27}x^3=x^2

    \Rightarrow 8x^3 - 27x^2=0

    \displaystyle \Rightarrow x = 0 or \displaystyle x = \tfrac{27}{8}

    \displaystyle x=0 clearly gives us problems with \log x. So we're left with the solution \displaystyle x = \tfrac{27}{8},\;y = \tfrac94.

    I haven't looked at (ii). I think you could begin in a similar way.

    Grandad
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  3. #3
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    LOG!!??

    Cant we solve it without using logarithms? Coz in this sum , my prof said that we are not allowed to use log!
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  4. #4
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    Hello Arka
    Quote Originally Posted by Arka View Post
    Cant we solve it without using logarithms? Coz in this sum , my prof said that we are not allowed to use log!
    Yes, you can; although it would have been helpful if you had said so at the beginning.

    I prefer to use logs, because it avoids handling fractional exponents, but here goes:
    y^3 = x^2

    \Rightarrow y = x^{\frac23}
    Substitute into the equation x^y = y^x:
    x^{(x^{\frac23})}=\big(x^{\frac23\big)^x}
    =x^{(\frac23x)}
    If we now compare the exponents on each side of the equation, we get:
    \Rightarrow x^{\frac23}=\frac23x
    Cube both sides:
    \Rightarrow x^2 = \frac{8}{27}x^3
    Then, as before, x = ...

    Grandad
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