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

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== Problem 9 ==
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== Problem ==
 
Let <math>x,</math> <math>y,</math> and <math>z</math> be positive real numbers that satisfy
 
Let <math>x,</math> <math>y,</math> and <math>z</math> be positive real numbers that satisfy
 
<cmath>2\log_{x}(2y) = 2\log_{2x}(4z) = \log_{2x^4}(8yz) \ne 0.</cmath>
 
<cmath>2\log_{x}(2y) = 2\log_{2x}(4z) = \log_{2x^4}(8yz) \ne 0.</cmath>
 
The value of <math>xy^5z</math> can be expressed in the form <math>\frac{1}{2^{p/q}},</math> where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q.</math>
 
The value of <math>xy^5z</math> can be expressed in the form <math>\frac{1}{2^{p/q}},</math> where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q.</math>
  
== Solution ==
+
== 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 make the problem as simple as possible let us assume with loss of generality that
+
Since there are only two dependent equations given and three unknowns, the three expressions given can equate to any common value (that isn't 0, of course), so to simplify the problem let us assume without loss of generality that
 
<cmath>2\log_{x}(2y) = 2\log_{2x}(4z) = \log_{2x^4}(8yz) = 2.</cmath>
 
<cmath>2\log_{x}(2y) = 2\log_{2x}(4z) = \log_{2x^4}(8yz) = 2.</cmath>
 
Then
 
Then
 
<cmath>
 
<cmath>
 
\begin{align*}
 
\begin{align*}
2\log_{x}(2y) = 2 &\longrightarrow x=2y\\
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2\log_{x}(2y) = 2 &\implies x=2y\\
2\log_{2x}(4z) = 2 &\longrightarrow 2x=4z\\
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2\log_{2x}(4z) = 2 &\implies 2x=4z\\
\log_{2x^4}(8yz) = 2 &\longrightarrow 4x^8 = 8yz
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\log_{2x^4}(8yz) = 2 &\implies 4x^8 = 8yz
 
\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
Solving these equations, we quickly see that <math>4x^8 = (2y)(4z) = x(2x) \longrightarrow x=2^{-1/6}</math> and then <math>y=z=2^{-1/6 - 1} = 2^{-7/6}.</math>
+
Solving these equations, we quickly see that <math>4x^8 = (2y)(4z) = x(2x) \rightarrow x=2^{-1/6}</math> and then <math>y=z=2^{-1/6 - 1} = 2^{-7/6}.</math>
Finally, our desired value is <math>2^{-1/6} \cdot (2^{-7/6})^5 \cdot 2^{-7/6} = 2^{-43/6}</math> and thus <math>m+n = 43 + 6 = \boxed{049.}</math>
+
Finally, our desired value is <math>2^{-1/6} \cdot (2^{-7/6})^5 \cdot 2^{-7/6} = 2^{-43/6}</math> and thus <math>p+q = 43 + 6 = \boxed{049.}</math>
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 +
==Solution 2==
 +
Notice that <math>2y\cdot 4z=8yz</math>, <math>2\log_2(2y)=\log_2\left(4y^2\right)</math> and <math>2\log_2(4z)=\log_2\left(16z^2\right)</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>
 +
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>
 +
<cmath>2x^4=\sqrt{x\cdot 2x}\implies 4x^8=2x^2\implies 2x^6=1\implies x=2^{-1/6}</cmath>
 +
We can plug <math>x</math> in to any of the two equal fractions aforementioned. So, without loss of generality:
 +
<cmath>\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)</cmath>
 +
<cmath>\implies \log\left(4y^2\right)\cdot \frac{5}{6}\log 2=\log\left(16z^2\right)\cdot \frac{-1}{6}\log 2</cmath>
 +
<cmath>\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</cmath>
 +
<cmath>\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}</cmath>
 +
 
 +
Thus <math>xy^5z=2^{-\frac{1}{6}-7}=2^{-\frac{43}{6}}</math> and <math>43+6=\boxed{049}</math>.
 +
 
 +
==Solution 3==
 +
Since we are given that <math>xy^5z = 2^{-p/q}</math>, we may assume that <math>x, y</math>, and <math>z</math> are all powers of two.  We shall thus let <math>x = 2^X</math>, <math>y = 2^Y</math>, and <math>z = 2^Z</math>.  Let <math>a = \log_{2^X}(2^{Y+1})</math>.  From this we get the system of equations: <cmath></cmath>
 +
<math>(1)</math><cmath>a = \log_{2^X}(2^{Y+1}) \Rightarrow aX = Y + 1</cmath>
 +
<math>(2)</math><cmath>a = \log_{2^{X + 1}}(2^{Z + 2}) \Rightarrow aX + a = Z + 2</cmath>
 +
<math>(3)</math><cmath>2a = \log_{2^{4X + 1}}(2^{Y + Z + 3}) \Rightarrow 8aX + 2a = Y + Z + 3</cmath>
 +
 
 +
Plugging equation <math>(1)</math> into equation <math>(2)</math> yields <math>Y + a = Z + 1</math>.  Plugging equation <math>(1)</math> into equation <math>(3)</math> and simplifying yields <math>7Y + 2a + 6 = Z + 1</math>, and substituting <math>Y + a</math> for <math>Z + 1</math> and simplifying yields <math>Y + 1 = \frac{-a}{6} </math>.  But <math>Y + 1 = aX</math>, so <math>aX =  \frac{-a}{6}</math>, so <math>X = \frac{-1}{6}</math>.
 +
 
 +
Knowing this, we may substitute <math>\frac{-1}{6}</math> for <math>X</math> in equations <math>(1)</math> and <math>(2)</math>, yielding <math>\frac{-a}{6} = Y + 1</math> and <math>\frac{5a}{6} = Z + 2</math>.  Thus, we have that <math>-5(Y + 1) = Z + 2 \rightarrow 5Y + Z = -7</math>.  We are looking for <math>xy^5z = 2^{X+ 5Y + Z}</math>. <math>X = \frac{-1}{6}</math> and <math>5Y + Z = -7</math>, so <math>xy^5z = 2^{-43/6} = \frac{1}{2^{43/6}}</math>.  The answer is <math>43+6=\boxed{049}</math>.
 +
 
 +
== Video Solution by Richard Rusczyk ==
 +
 
 +
https://artofproblemsolving.com/videos/amc/2012aimei/348
 +
 
 +
~ dolphin7
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2012|n=I|num-b=8|num-a=10}}
 
{{AIME box|year=2012|n=I|num-b=8|num-a=10}}
 +
 +
[[Category:Intermediate Algebra Problems]]
 +
{{MAA Notice}}

Revision as of 20:21, 24 January 2021

Problem

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 (that isn't 0, of course), 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 $xy^5z=2^{-\frac{1}{6}-7}=2^{-\frac{43}{6}}$ and $43+6=\boxed{049}$.

Solution 3

Since we are given that $xy^5z = 2^{-p/q}$, we may assume that $x, y$, and $z$ are all powers of two. We shall thus let $x = 2^X$, $y = 2^Y$, and $z = 2^Z$. Let $a = \log_{2^X}(2^{Y+1})$. From this we get the system of equations: \[\] $(1)$\[a = \log_{2^X}(2^{Y+1}) \Rightarrow aX = Y + 1\] $(2)$\[a = \log_{2^{X + 1}}(2^{Z + 2}) \Rightarrow aX + a = Z + 2\] $(3)$\[2a = \log_{2^{4X + 1}}(2^{Y + Z + 3}) \Rightarrow 8aX + 2a = Y + Z + 3\]

Plugging equation $(1)$ into equation $(2)$ yields $Y + a = Z + 1$. Plugging equation $(1)$ into equation $(3)$ and simplifying yields $7Y + 2a + 6 = Z + 1$, and substituting $Y + a$ for $Z + 1$ and simplifying yields $Y + 1 = \frac{-a}{6}$. But $Y + 1 = aX$, so $aX =  \frac{-a}{6}$, so $X = \frac{-1}{6}$.

Knowing this, we may substitute $\frac{-1}{6}$ for $X$ in equations $(1)$ and $(2)$, yielding $\frac{-a}{6} = Y + 1$ and $\frac{5a}{6} = Z + 2$. Thus, we have that $-5(Y + 1) = Z + 2 \rightarrow 5Y + Z = -7$. We are looking for $xy^5z = 2^{X+ 5Y + Z}$. $X = \frac{-1}{6}$ and $5Y + Z = -7$, so $xy^5z = 2^{-43/6} = \frac{1}{2^{43/6}}$. The answer is $43+6=\boxed{049}$.

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2012aimei/348

~ dolphin7

See also

2012 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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