Difference between revisions of "2012 AIME I Problems/Problem 9"
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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> | + | 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> |
== See also == | == See also == | ||
Revision as of 19:07, 14 February 2015
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
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 
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. 
