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May 25th, 2014, 06:25 PM   #1
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Math Competition Problem

How should I approach this problem?

Screenshot 2014-05-25 at 10.21.11 PM.jpg

12. $\,$ If $\log_2x$ and $\log_2y$ are distinct positive integers and $\log_x2+\log_y2=0.5,\, xy=$
A. 64 $\quad$ B. 128 $\quad$ C. 256 $\quad$ D. 512 $\quad$ E. 1024

Last edited by skipjack; June 20th, 2014 at 03:44 AM.
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May 25th, 2014, 06:30 PM   #2
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I would start with $$\log_a{b} = \frac{\log b}{\log a}$$
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May 25th, 2014, 07:01 PM   #3
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I thought of that. It lead me to a dead end
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May 25th, 2014, 07:33 PM   #4
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xy = 512. (1/3 + 1/6 = 1/2).
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May 25th, 2014, 07:36 PM   #5
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Greg, how were you able to deduce that the x and y values were 1/3 and 1/6?
Did you do anything in particular or you just test a few numbers?
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May 25th, 2014, 08:08 PM   #6
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Hello, Orlando895!

Quote:
12. If $\displaystyle \log_2x $ and $\displaystyle \log_2y$ are distinct positive integers,
$\displaystyle \quad\;\;\;$and $\displaystyle \log_x2 + \log_y2 \,=\,\tfrac{1}{2}$, find $\displaystyle xy.$
$\displaystyle \qquad\quad (A)\;64\qquad (B)\;128 \qquad (C)\;256 \qquad (D)\;512\qquad (E)\;1024$

We have: $\displaystyle \;\log_x2 + \log_y2 \:=\:\frac{1}{2} \quad\Rightarrow\quad \frac{1}{\log_2x} + \frac{1}{\log_2y} \:=\:\frac{1}{2}$

$\displaystyle \quad \frac{\log_2x+\log_2y}{\log_2x\log_2y} \:=\:\frac{1}{2} \quad\Rightarrow\quad 2\log_2x + 2\log_2y \:=\:\log_2x\!\cdot\!\log_2y $

$\displaystyle \quad \log_2x\!\cdot\!\log_2y - 2\log_2y \;=\;2\log_2x \quad\Rightarrow\quad (\log_2x - 2)\log_2y \:=\:2\log_2x$

$\displaystyle \quad \log_2y \:=\:\frac{2\log_2x}{\log_2x-2}$

We have the form: $\displaystyle \:B \;=\;\frac{2A}{A-2}$

Since $\displaystyle A$ and $\displaystyle B$ are positive integers,
$\displaystyle \quad$the only solutions for $\displaystyle A$ are: $\displaystyle \,A = 3,4,6.$

Then the solutions for $\displaystyle B$ are: $\displaystyle \;B \,=\,6,4,3$

Since $\displaystyle a$ and $\displaystyle b$ are distinct integers,
$\displaystyle \quad$ the solutions are: $\displaystyle \:(\log_2x,\,\log_2y) \;=\;(3,6),\;(6,3)$

One solution is: $\displaystyle \:\begin{Bmatrix}\log_2x = 3 & \Rightarrow & x = 2^3 = 8 \\ \log_2y = 6 & \Rightarrow & y = 2^6 =64 \end{Bmatrix}$

Therefore: $\displaystyle \:xy \:=\:8\cdot64 \:=\:512\;\text{ . . . Answer (D)}$
Thanks from Orlando895

Last edited by skipjack; June 20th, 2014 at 03:48 AM.
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May 25th, 2014, 11:06 PM   #7
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Soroban: Wow, very nice!!! Thank you so much. Your steps are all very clear. Thanks a ton
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May 27th, 2014, 08:51 PM   #8
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Quote:
Originally Posted by Orlando895 View Post
Greg, how were you able to deduce that the x and y values were 1/3 and 1/6?
Did you do anything in particular or you just test a few numbers?
Knowing that 1/3 + 1/6 = 1/2 helped. In fact, 1/a + 1/b = 1/2 has as its only solutions
those that satisfy 2n ≡ 0 (mod n - 2), that is (a, b) = (3, 6), (4, 4) and (6, 3).
As $\log_2x$ and $\log_2y$ are distinct positive integers the result follows.

$\displaystyle \left(\log_2x\right)^{-1}=\log_{x}2$
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