Difference between revisions of "2012 AIME I Problems/Problem 9"
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<cmath> | <cmath> | ||
\begin{align*} | \begin{align*} | ||
− | 2\log_{x}(2y) = 2 &\ | + | 2\log_{x}(2y) = 2 &\rightarrow x=2y\\ |
− | 2\log_{2x}(4z) = 2 &\ | + | 2\log_{2x}(4z) = 2 &\rightarrow 2x=4z\\ |
− | \log_{2x^4}(8yz) = 2 &\ | + | \log_{2x^4}(8yz) = 2 &\rightarrow 4x^8 = 8yz |
\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
− | Solving these equations, we quickly see that <math>4x^8 = (2y)(4z) = x(2x) \ | + | 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>m+n = 43 + 6 = \boxed{049.}</math> | ||
== 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}} |
Revision as of 02:22, 17 March 2012
Problem 9
Let
and
be positive real numbers that satisfy
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 make the problem as simple as possible let us assume with loss of generality that
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 |