Logaritmi
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Adriana Nistor
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Logaritmi
Daca \( a,b,c\in (1,\inf) \), \( abc=8 \) si \( \log_{a}{(a-1)}+\log_{b}{(b-1)}+\log_{c}{(c-1)}=0 \), atunci aflati numerele reale \( a,b,c \).
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Marius Mainea
- Gauss
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a=b=c=2 (se foloseste AM-GM)
\( 0=\sum\log_a(a-1)\cdot1\le\sum\log_a(\frac{a-1+1}{2})^2=6-\sum\frac{1}{\log_4{a}}\le 6-\frac{9}{\sum\log_4a}=0 \)
E bine asa ?
\( 0=\sum\log_a(a-1)\cdot1\le\sum\log_a(\frac{a-1+1}{2})^2=6-\sum\frac{1}{\log_4{a}}\le 6-\frac{9}{\sum\log_4a}=0 \)
E bine asa ?
Last edited by Marius Mainea on Tue Sep 22, 2009 10:48 am, edited 2 times in total.
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Adriana Nistor
- Pitagora
- Posts: 82
- Joined: Thu Aug 07, 2008 10:07 pm
- Location: Drobeta Turnu Severin, Mehedinti