Math Competition Problem

Nov 2013
47
4
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
 
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v8archie

Math Team
Dec 2013
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Colombia
I would start with $$\log_a{b} = \frac{\log b}{\log a}$$
 

greg1313

Forum Staff
Oct 2008
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London, Ontario, Canada - The Forest City
xy = 512. (1/3 + 1/6 = 1/2).
 
Nov 2013
47
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?
 

soroban

Math Team
Dec 2006
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408
Lexington, MA
Hello, Orlando895!

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: \(\displaystyle \:xy \:=\:8\cdot64 \:=\:512\;\text{ . . . Answer (D)}\)
 
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Nov 2013
47
4
Soroban: Wow, very nice!!! Thank you so much. Your steps are all very clear. Thanks a ton :D
 

greg1313

Forum Staff
Oct 2008
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London, Ontario, Canada - The Forest City
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.

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