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

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(→‎Solution 2)
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\log\frac{x}{z}(1-\log y) &= 0 \\
 
\log\frac{x}{z}(1-\log y) &= 0 \\
 
\end{align*}</cmath>
 
\end{align*}</cmath>
−
If <math>\log\frac{x}{z}=0</math> then <math>\frac{x}{z}=1\Longrightarrow x=z</math>.
+
If <math>\log\frac{x}{z}=0</math> then <math>\frac{x}{z}=1\Longrightarrow x=z</math>. Substituting into the third equation gets
 +
<cmath>\begin{align*}
 +
\log x^2-(\log x)(\log x) &= 0 \\
 +
\log x^2-\log x^x &= 0 \\
 +
\log x^{2-x} &= 0 \\
 +
x^{2-x} &= 1 \\
 +
\end{align*}</cmath>
 +
Thus either <math>x=1</math> or <math>2-x=0\Longrightarrow x=2</math>. (Note that here <math>x\neq-1</math> since logarithm isn't defined for negative number.)
 +
 
 +
If
  
 
== See also ==
 
== See also ==

Revision as of 22:36, 6 December 2019

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}(zx) - (\log_{10}z)(\log_{10}x) & = & 0 \\ \end{eqnarray*}

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

Solution

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 - \log 2000\\ -\log y \log z + \log y + \log z - 1 &=& - \log 2\\ -\log x \log z + \log x + \log z - 1 &=& -1\\ \end{eqnarray*}

Let $a,b,c$ be $\log x, \log y, \log z$ respectively. Using SFFT, the above equations become (*)

\begin{eqnarray*}(a - 1)(b - 1) &=& \log 2 \\ (b-1)(c-1) &=& \log 2 \\ (a-1)(c-1) &=& 1 \end{eqnarray*}

From here, multiplying the three equations gives

\begin{eqnarray*}(a-1)^2(b-1)^2(c-1)^2 &=& (\log 2)^2\\ (a-1)(b-1)(c-1) &=& \pm\log 2\end{eqnarray*}

Dividing the third equation of (*) from this equation, $b-1 = \log y - 1 = \pm\log 2 \Longrightarrow \log y = \pm \log 2 + 1$. This gives $y_1 = 20, y_2 = 5$, and the answer is $y_1 + y_2 = \boxed{025}$.

Solution 2

Subtracting the second equation from the first equation yields \begin{align*} \log 2000xy-\log 2yz-((\log x)(\log y)-(\log y)(\log z)) &= 3 \\ \log\frac{2000xy}{2yz}-\log y(\log x-\log z) &= 3 \\ \log1000+\log\frac{x}{z}-\log y(\log\frac{x}{z}) &= 3 \\ 3+\log\frac{x}{z}-\log y(\log\frac{x}{z}) &= 3 \\ \log\frac{x}{z}(1-\log y) &= 0 \\ \end{align*} If $\log\frac{x}{z}=0$ then $\frac{x}{z}=1\Longrightarrow x=z$. Substituting into the third equation gets \begin{align*} \log x^2-(\log x)(\log x) &= 0 \\ \log x^2-\log x^x &= 0 \\ \log x^{2-x} &= 0 \\ x^{2-x} &= 1 \\ \end{align*} Thus either $x=1$ or $2-x=0\Longrightarrow x=2$. (Note that here $x\neq-1$ since logarithm isn't defined for negative number.)

If

See also

2000 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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