Difference between revisions of "2012 AIME I Problems/Problem 9"

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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>
 
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>
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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>p+q = 43 + 6 = \boxed{049.}</math>
  
 
== See also ==
 
== See also ==

Revision as of 19:07, 14 February 2015

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 simplify the problem let us assume without 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 &\implies x=2y\\ 2\log_{2x}(4z) = 2 &\implies 2x=4z\\ \log_{2x^4}(8yz) = 2 &\implies 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 $p+q = 43 + 6 = \boxed{049.}$

See also

2012 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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All AIME Problems and Solutions

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