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 May 25th, 2014, 06:25 PM #1 Member   Joined: Nov 2013 Posts: 47 Thanks: 4 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. May 25th, 2014, 06:30 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,683 Thanks: 2664 Math Focus: Mainly analysis and algebra I would start with $$\log_a{b} = \frac{\log b}{\log a}$$ May 25th, 2014, 07:01 PM #3 Member   Joined: Nov 2013 Posts: 47 Thanks: 4 I thought of that. It lead me to a dead end  May 25th, 2014, 07:33 PM #4 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond xy = 512. (1/3 + 1/6 = 1/2). May 25th, 2014, 07:36 PM #5 Member   Joined: Nov 2013 Posts: 47 Thanks: 4 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? 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: $\displaystyle \:xy \:=\:8\cdot64 \:=\:512\;\text{ . . . Answer (D)}$

Last edited by skipjack; June 20th, 2014 at 03:48 AM. May 25th, 2014, 11:06 PM #7 Member   Joined: Nov 2013 Posts: 47 Thanks: 4 Soroban: Wow, very nice!!! Thank you so much. Your steps are all very clear. Thanks a ton  May 27th, 2014, 08:51 PM   #8
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Quote:
 Originally Posted by Orlando895 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.

$\displaystyle \left(\log_2x\right)^{-1}=\log_{x}2$ Tags competition, math, problem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Orlando895 Math Events 3 May 24th, 2014 07:56 AM Orlando895 Trigonometry 4 April 1st, 2014 10:54 AM icemanfan Algebra 3 March 15th, 2012 06:45 PM PaidGrade New Users 0 March 14th, 2012 11:50 AM asddsa Algebra 0 November 16th, 2011 03:06 PM

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