Difference between revisions of "2012 AIME I Problems/Problem 9"
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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 simplify the problem let us assume without 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, 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> | ||
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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> | 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>p+q = 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> | ||
| + | |||
| + | ==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 an arithmetic 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)</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^10z^2=1\implies y^10z^2=\frac{1}{16384}\implies y^5z=\frac{1}{128}</cmath> | ||
| + | |||
| + | Thus <math>xy^5z=2^{-1/6-7}=2^{-43/6}</math> and <math>43+6=\boxed{049}</math>. | ||
== See also == | == See also == | ||
Revision as of 20:03, 21 February 2016
Contents
Problem 9
Let 
and 
be positive real numbers that satisfy
![\[2\log_{x}(2y) = 2\log_{2x}(4z) = \log_{2x^4}(8yz) \ne 0.\]](http://latex.artofproblemsolving.com/0/0/4/0040698f1dfdbbb5d2fe34839edc567e8818e9ae.png)
The value of 
can be expressed in the form 
where 
and 
are relatively prime positive integers. Find 
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
![\[2\log_{x}(2y) = 2\log_{2x}(4z) = \log_{2x^4}(8yz) = 2.\]](http://latex.artofproblemsolving.com/d/3/e/d3e7b810a6d2b978439468b4ddc05ea235f7a990.png)
Then
Solving these equations, we quickly see that 
and then 
Finally, our desired value is 
and thus 
Solution 2
Notice that 
, 
and 
.
From this, we see that 
is the geometric mean of 
and 
. So, for constant 
:
![\[\frac{\log 4y^2}{\log x}=\frac{\log 8yz}{\log 2x^4}=\frac{\log 16z^2}{\log 2x}=C\]](http://latex.artofproblemsolving.com/2/3/f/23f39de85e1f5f39b0ad00fc3dec565c1fa27e81.png)
Since 
are in an arithmetic progression, so is 
.
Therefore, 
is the geometric mean of 
and 
![\[2x^4=\sqrt{x\cdot 2x}\implies 4x^8=2x^2\implies 2x^6=1\implies x=2^{-1/6}\]](http://latex.artofproblemsolving.com/6/a/a/6aa1d19c23f1d7a482fca4d2460fcd349ff5f9e1.png)
We can plug in to any of the two equal fractions aforementioned. So, without loss of generality:
\[\frac{\log 4y^2}{\log x}=\frac{\log 16z^2}{\log 2x}\implies \log\left(4y^2\right)\log\left(2x\right)=\log\left(16z^2\right)\log\left(x)\] (Error compiling LaTeX. Unknown error_msg)
![\[\implies \log\left(4y^2\right)\cdot \frac{5}{6}\log 2=\log\left(16z^2\right)\cdot \frac{-1}{6}\log 2\]](http://latex.artofproblemsolving.com/b/6/5/b65bdb7a5d6d9e2b9ad1a72414128c53e88e0555.png)
![\[\implies 5\log\left(4y^2\right)=-\log\left(16z^2\right)\implies 5\log\left(4y^2\right)+\log\left(16z^2\right)=0\]](http://latex.artofproblemsolving.com/b/c/f/bcfd0aba1264651023e6d4a00201d0f1b6eba82c.png)
![\[\implies \left(4y^2\right)^5\cdot 16z^2=1\implies 16384y^10z^2=1\implies y^10z^2=\frac{1}{16384}\implies y^5z=\frac{1}{128}\]](http://latex.artofproblemsolving.com/f/e/a/feafbd394fc18938d611063fb790f70313da260c.png)
Thus 
and 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
