# Thread: System of Equations - Logarithms

1. ## System of Equations - Logarithms

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

$\displaystyle \log (2xy)=\log x \log y$
$\displaystyle \log(yz)=\log y \log z$
$\displaystyle \log(2zx)=\log z^2 \log x$

2. Originally Posted by Winding Function
Solve the following system of equations in positive real numbers (the logarithm is in base 10):

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

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

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

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