2000 AIME I Problems/Problem 9

Problem

The system of equations

$\begin{eqnarray*}\log_{10}(2000xy) - (\log_{10}x)(\log_{10}y) & = & 4 \\

\log_{10}(2yz) - (\log_{10}y)(\log_{10}z) & = & 1 \\ \log_{10}(2000zx) - (\log_{10}z)(\log_{10}x) & = & 0 \\

\end{eqnarray*}$ (Error compiling LaTeX. Unknown error_msg)

has two solutions $(x_{1},y_{1},z_{1})$ and $(x_{2},y_{2},z_{2})$ . Find $y_{1} + y_{2}$ .

Will finish this problem soon. Azjps (talk) 18:12, 31 December 2007 (EST)

Since $\log ab = \log a + \log b$ , we can reduce the equations to a more recognizable form:

$\begin{eqnarray*}- (\log x)(\log y) + \log x + \log y - 1 & = & 3 - 3\log 2 \\ etc &=& etc \end{eqnarray*}$

Let $x_1, y_1, z_1$ be $\log x, \log y, \log z$ respectively. Using SFFT, it becomes

$\begin{eqnarray*}(x_1 - 1)(y_1 - 1) &=& 3\log2 - 3\\ etc &=& etc \end{eqnarray*}$

Multiplying the three equations gives

$\begin{eqnarray*}(x_1-1)^2(y_1-1)^2(z_1-1)^2 &=& something\\ (x_1-1)(y_1-1)(z_1-1) &=& \sqrt{something}\end{eqnarray*}$

We can now divide each of the previous equations from this equation to get $y_1 - 1 = something$ , so the answer is $\boxed{answer}$ .

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