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Difference between revisions of "2012 AIME I Problems/Problem 9"
m (→‎Solution)
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<cmath>
 
<cmath>
 
\begin{align*}
 
\begin{align*}
−
2\log_{x}(2y) = 2 &\longrightarrow x=2y\\
+
2\log_{x}(2y) = 2 &\rightarrow x=2y\\
−
2\log_{2x}(4z) = 2 &\longrightarrow 2x=4z\\
+
2\log_{2x}(4z) = 2 &\rightarrow 2x=4z\\
−
\log_{2x^4}(8yz) = 2 &\longrightarrow 4x^8 = 8yz
+
\log_{2x^4}(8yz) = 2 &\rightarrow 4x^8 = 8yz
 
\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
−
Solving these equations, we quickly see that <math>4x^8 = (2y)(4z) = x(2x) \longrightarrow x=2^{-1/6}</math> and then <math>y=z=2^{-1/6 - 1} = 2^{-7/6}.</math>
+
Solving these equations, we quickly see that <math>4x^8 = (2y)(4z) = x(2x) \rightarrow x=2^{-1/6}</math> and then <math>y=z=2^{-1/6 - 1} = 2^{-7/6}.</math>
 
Finally, our desired value is <math>2^{-1/6} \cdot (2^{-7/6})^5 \cdot 2^{-7/6} = 2^{-43/6}</math> and thus <math>m+n = 43 + 6 = \boxed{049.}</math>
 
Finally, our desired value is <math>2^{-1/6} \cdot (2^{-7/6})^5 \cdot 2^{-7/6} = 2^{-43/6}</math> and thus <math>m+n = 43 + 6 = \boxed{049.}</math>
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2012|n=I|num-b=8|num-a=10}}
 
{{AIME box|year=2012|n=I|num-b=8|num-a=10}}

Revision as of 02:22, 17 March 2012

Problem 9

Let $x,$ $y,$ and $z$ be positive real numbers that satisfy \[2\log_{x}(2y) = 2\log_{2x}(4z) = \log_{2x^4}(8yz) \ne 0.\] The value of $xy^5z$ can be expressed in the form $\frac{1}{2^{p/q}},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$

Solution

Since there are only two dependent equations given and three unknowns, the three expressions given can equate to any common value, so to make the problem as simple as possible let us assume with loss of generality that \[2\log_{x}(2y) = 2\log_{2x}(4z) = \log_{2x^4}(8yz) = 2.\] Then \begin{align*} 2\log_{x}(2y) = 2 &\rightarrow x=2y\\ 2\log_{2x}(4z) = 2 &\rightarrow 2x=4z\\ \log_{2x^4}(8yz) = 2 &\rightarrow 4x^8 = 8yz \end{align*} Solving these equations, we quickly see that $4x^8 = (2y)(4z) = x(2x) \rightarrow x=2^{-1/6}$ and then $y=z=2^{-1/6 - 1} = 2^{-7/6}.$ Finally, our desired value is $2^{-1/6} \cdot (2^{-7/6})^5 \cdot 2^{-7/6} = 2^{-43/6}$ and thus $m+n = 43 + 6 = \boxed{049.}$

See also

2012 AIME I (Problems • Answer Key • Resources)
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Problem 8 		Followed by
Problem 10
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All AIME Problems and Solutions
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