Difference between revisions of "2000 AIME I Problems/Problem 9"

Revision as of 19:12, 31 December 2007

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 $\boxed{answer}$ .

@@ Line 9: / Line 9: @@
 == Solution ==
-{{solution}}
+:''Will finish this problem soon. <font style="font-family:Georgia,sans-serif">[[User:Azjps|Azjps]] ([[User talk:Azjps|<font color="green">talk</font>]])</font> 18:12, 31 December 2007 (EST)''
+Since <math>\log ab = \log a + \log b</math>, we can reduce the equations to a more recognizable form:
+<cmath>\begin{eqnarray*}- (\log x)(\log y) + \log x + \log y - 1 & = & 3 - 3\log 2 \\
+etc &=& etc
+\end{eqnarray*}</cmath>
+Let <math>x_1, y_1, z_1</math> be <math>\log x, \log y, \log z</math> respectively. Using [[SFFT]], it becomes
+<cmath>\begin{eqnarray*}(x_1 - 1)(y_1 - 1) &=& 3\log2 - 3\\
+etc &=& etc
+\end{eqnarray*}</cmath>
+Multiplying the three equations gives
+<cmath>\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*}</cmath>
+We can now divide each of the previous equations from this equation to get <math>y_1 - 1 = something</math>, so the answer is <math>\boxed{answer}</math>.
 == See also ==
 {{AIME box|year=2000|n=I|num-b=8|num-a=10}}
+[[Category:Intermediate Algebra Problems]]

2000 AIME I (Problems • Answer Key • Resources)
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