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Thread: System of Equations - Logarithms

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    System of Equations - Logarithms

    Solve the following system of equations in positive real numbers (the logarithm is in base 10):

    \log (2xy)=\log x \log y
    \log(yz)=\log y \log z
    \log(2zx)=\log z^2 \log x
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    Quote Originally Posted by Winding Function View Post
    Solve the following system of equations in positive real numbers (the logarithm is in base 10):

    \log (2xy)=\log x \log y
    \log(yz)=\log y \log z
    \log(2zx)=\log z^2 \log x
    Let a=\log x,\;b=\log y,\;c=\log z,\;m=\log 2. Then the equations become

    m+a+b=ab,\quad b+c=bc,\quad m+c+a = 2ca.

    The first two equations can be written (a-1)(b-1)=m+1 and (b-1)(c-1)=1, from which it follows that a-1 = (m+1)(c-1) and so a = (m+1)c-m. Substitute that into the third equation, and you find that c = \frac{3m+2}{2m+2}, from which it follows that a = \frac{m+2}2 and b = \frac{3m+2}m.

    So 2a = m+2, or 2\log x = \log 2 + \log 100, from which x^2 = 200, x = 10\sqrt2. The expressions for y and z are similar but more awkward, because they will involve raising numbers to irrational powers.
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