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 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 $m+n = 43 + 6 = \boxed{049.}$

2012 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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2012 AIME I Problems/Problem 9

Problem 9

Solution

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