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

Revision as of 21:23, 1 April 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 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 $m+n = 43 + 6 = \boxed{049.}$

@@ Line 20: / Line 20: @@
 == See also ==
 {{AIME box|year=2012|n=I|num-b=8|num-a=10}}
+[[Category:Intermediate Algebra Problems]]

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 21:23, 1 April 2012

Problem 9

Solution

See also