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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 20140525 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: 7,344 Thanks: 2466 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: 7,842 Thanks: 1068 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  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 407  Hello, Orlando895! Quote:
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 $ $\displaystyle \quad \log_2x\!\cdot\!\log_2y  2\log_2y \;=\;2\log_2x \quad\Rightarrow\quad (\log_2x  2)\log_2y \:=\:2\log_2x$ $\displaystyle \quad \log_2y \:=\:\frac{2\log_2x}{\log_2x2}$ We have the form: $\displaystyle \:B \;=\;\frac{2A}{A2}$ 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  
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,842 Thanks: 1068 Math Focus: Elementary mathematics and beyond  Quote:
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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