System of logarithmic equations

Question

$$\log (2000xy)-\log x\log y=4$$$$\log(2yz)-\log y\log z=1$$$$\log(zx)-\log z\log x=0$$ The base is 10 everywhere. I tried opening the log with the sum formulae and then manipulating, but I got stuck. Conventional methods don't seem to help. Is there some trick? What am I missing?

Answer 1

There are definitely tricks. I have no idea whether this solution is the intended, but anyway, here goes.

Lets start with the last equation and tweak it a bit.$$\begin{align*} \log(zx)-\log z\log x &=0 \\ 1 - \log z -\log x +\log z\log x &= 1\\ (1- \log x)(1- \log z) &= 1.\end{align*}$$

I would then let $X = 1- \log x$, $Y = 1- \log y$ and $Z = 1- \log z$ and do the same sort of thing to every equation. You'll get expressions for $XY$, $YZ$ and $XZ$ as constants. Doing a division of one equation by another, then multiplying that by the third equation, and you should be getting things you can solve easily.