# Math Competition Problem

#### Orlando895

How should I approach this problem?

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 a moderator:

#### v8archie

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

#### greg1313

Forum Staff
xy = 512. (1/3 + 1/6 = 1/2).

#### 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?

#### soroban

Math Team
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: $$\displaystyle \:xy \:=\:8\cdot64 \:=\:512\;\text{ . . . Answer (D)}$$

Last edited by a moderator:
1 person

#### Orlando895

Soroban: Wow, very nice!!! Thank you so much. Your steps are all very clear. Thanks a ton

#### greg1313

Forum Staff
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 â‰¡ 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.

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