2012 AIME I Problems/Problem 9

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

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