Mathematical and Physical Journal
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Problem C. 1033. (April 2010)

C. 1033. Solve the equation \(\displaystyle \log_{2010}\, (2009\, x)=\log_{2009}\, (2010\, x)\).

Suggested by J. Pataki, Budapest

(5 pont)

Deadline expired on May 10, 2010.


Sorry, the solution is available only in Hungarian. Google translation

Megoldás. A feladat értelmezési tartománya: \(\displaystyle x>0\). Áttérve 2009-es alapra:

\(\displaystyle \frac{\log_{2009} (2009x)}{\log_{2009} 2010}=\log_{2009} (2010x). \)

Szorozhatunk a nem 0 nevezővel:

\(\displaystyle \log_{2009} (2009x) =\log_{2009} 2010\cdot \log_{2009} (2010x). \)

Alkalmazzuk a szorzat logaritmusára ismert összefüggést:

\(\displaystyle \log_{2009} 2009+\log_{2009} x =\log_{2009} 2010\cdot (\log_{2009} 2010+\log_{2009} x),\)

\(\displaystyle 1+\log_{2009} x =\log_{2009}^2 2010+\log_{2009} 2010\cdot \log_{2009} x,\)

\(\displaystyle \log_{2009} x =-\frac{\log_{2009}^2 2010-1}{\log_{2009} 2010-1},\)

\(\displaystyle \log_{2009} x =-\log_{2009} 2010-1, \)

\(\displaystyle \log_{2009} x =\log_{2009} 2010^{-1}-\log_{2009} 2009, \)

\(\displaystyle \log_{2009} x =\log_{2009} \frac{1}{2009\cdot 2010}.\)

A \(\displaystyle \log_{2009} x\) függvény szigorú monotonitása miatt ez akkor és csak akkor teljesül, ha

\(\displaystyle x=\frac{1}{2009\cdot 2010}, \)

ami eleget tesz a kikötésnek. (Ellenőrzéssel látható, hogy jól számoltunk.)

Blóz Gizella Evelin (Paks, Vak Bottyán Gimn., 11. évf.)


Statistics:

148 students sent a solution.
5 points:59 students.
4 points:42 students.
3 points:28 students.
2 points:10 students.
1 point:2 students.
0 point:7 students.

Problems in Mathematics of KöMaL, April 2010