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

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From this, we see that <math>8yz</math> is the geometric mean of <math>4y^2</math> and <math>16z^2</math>. So, for constant <math>C\ne 0</math>:
 
From this, we see that <math>8yz</math> is the geometric mean of <math>4y^2</math> and <math>16z^2</math>. So, for constant <math>C\ne 0</math>:
 
<cmath>\frac{\log 4y^2}{\log x}=\frac{\log 8yz}{\log 2x^4}=\frac{\log 16z^2}{\log 2x}=C</cmath>
 
<cmath>\frac{\log 4y^2}{\log x}=\frac{\log 8yz}{\log 2x^4}=\frac{\log 16z^2}{\log 2x}=C</cmath>
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Since <math>\log 4y^2,\log 8yz,\log 16z^2</math> are in an arithmetic progression, so is <math>\log x,\log 2x^4,\log 2x</math>.
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Since <math>\log 4y^2,\log 8yz,\log 16z^2</math> are in a geometric progression, so is <math>\log x,\log 2x^4,\log 2x</math>.
  
 
Therefore, <math>2x^4</math> is the geometric mean of <math>x</math> and <math>2x</math>
 
Therefore, <math>2x^4</math> is the geometric mean of <math>x</math> and <math>2x</math>

Revision as of 00:26, 21 August 2016

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 1

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 $p+q = 43 + 6 = \boxed{049.}$

Solution 2

Notice that $2y\cdot 4z=8yz$, $2\log_2(2y)=\log_2\left(4y^2\right)$ and $2\log_2(4z)=\log_2\left(16z^2\right)$.

From this, we see that $8yz$ is the geometric mean of $4y^2$ and $16z^2$. So, for constant $C\ne 0$: \[\frac{\log 4y^2}{\log x}=\frac{\log 8yz}{\log 2x^4}=\frac{\log 16z^2}{\log 2x}=C\] Since $\log 4y^2,\log 8yz,\log 16z^2$ are in a geometric progression, so is $\log x,\log 2x^4,\log 2x$.

Therefore, $2x^4$ is the geometric mean of $x$ and $2x$ \[2x^4=\sqrt{x\cdot 2x}\implies 4x^8=2x^2\implies 2x^6=1\implies x=2^{-1/6}\] We can plug $x$ in to any of the two equal fractions aforementioned. So, without loss of generality: \[\frac{\log 4y^2}{\log x}=\frac{\log 16z^2}{\log 2x}\implies \log\left(4y^2\right)\log\left(2x\right)=\log\left(16z^2\right)\log\left(x\right)\] \[\implies \log\left(4y^2\right)\cdot \frac{5}{6}\log 2=\log\left(16z^2\right)\cdot \frac{-1}{6}\log 2\] \[\implies 5\log\left(4y^2\right)=-\log\left(16z^2\right)\implies 5\log\left(4y^2\right)+\log\left(16z^2\right)=0\] \[\implies \left(4y^2\right)^5\cdot 16z^2=1\implies 16384y^{10}z^2=1\implies y^{10}z^2=\frac{1}{16384}\implies y^5z=\frac{1}{128}\]

Thus $xy^5z=2^{-1/6-7}=2^{-943/6}$ and $943+6=\boxed{949}$.

See also

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

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