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

Revision as of 00:02, 9 February 2014

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 $\begin{align*} 2\log_{x}(2y) = 2 &\implies x=2y\\ 2\log_{2x}(4z) = 2 &\implies 2x=4z\\ \log_{2x^4}(8yz) = 2 &\implies 4x^8 = 8yz \end{align*}$ 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 $m+n = 43 + 6 = \boxed{049.}$

2012 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

Page

Toolbox

Search

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

Revision as of 00:02, 9 February 2014

Problem 9

Solution

See also

@@ Line 10: / Line 10: @@
 <cmath>
 \begin{align*}
-\log_{x}(2y) = 2 &\rightarrow x=2y\\
+\log_{x}(2y) = 2 &\implies x=2y\\
-\log_{2x}(4z) = 2 &\rightarrow 2x=4z\\
+\log_{2x}(4z) = 2 &\implies 2x=4z\\
-\log_{2x^4}(8yz) = 2 &\rightarrow 4x^8 = 8yz
+\log_{2x^4}(8yz) = 2 &\implies 4x^8 = 8yz
 \end{align*}
 </cmath>