Rezolvati in Z ecuatia:
\( 9^x+4^x+2^x=8^x+6^x+1 \)
G.M. 6/2009
Ecuatie in Z
Moderators: Laurian Filip, Filip Chindea, maky, Cosmin Pohoata
-
Marius Mainea
- Gauss
- Posts: 1077
- Joined: Mon May 26, 2008 2:12 pm
- Location: Gaesti (Dambovita)
-
Mateescu Constantin
- Newton
- Posts: 307
- Joined: Tue Apr 21, 2009 8:17 am
- Location: Pitesti
Pentru \( x\ <\ 0\ \Longrightarrow\ 9^x\ \le\ \frac 19 \) , \( 4^x\ \le\ \frac 14 \) , \( 2^x\ \le\ \frac 12 \) , deci \( RHS\ \le\ \frac{31}{36}\ <\ 1 \) si \( LHS\ >\ 1 \), contradictie.
\( x=0 \) , \( x=1 \) , \( x=2 \) verifica in mod evident enuntul . Sa demonstram ca pentru \( x\ge 3 \) nu avem solutii.
\( 8^x\ +\ 6^x=8^{x-3}\ \cdot\ 8^3\ +\ 6^{x-3}\ \cdot\ 6^3\ \le\ 9^{x-3}\ \cdot\ 8^3\ +\ 9^{x-3}\ \cdot\ 6^3=9^x\ \frac{8^3}{9^3}\ +\ 9^x\ \frac{6^3}{9^3}=9^x\ \frac{728}{729}\ <\ 9^x \)
Deci \( 9^x\ >\ 8^x\ +\ 6^x, \ x\ge 3 \) si cum \( 4^x\ +\ 2^x\ >\ 1 \ \Longrightarrow\ RHS\ >\ LHS\ ,\ \ x\ge 3 \) . In concluzie \( x\in\{\ 0\ ,\ 1\ ,\ 2\ \} \) .
\( x=0 \) , \( x=1 \) , \( x=2 \) verifica in mod evident enuntul . Sa demonstram ca pentru \( x\ge 3 \) nu avem solutii.
\( 8^x\ +\ 6^x=8^{x-3}\ \cdot\ 8^3\ +\ 6^{x-3}\ \cdot\ 6^3\ \le\ 9^{x-3}\ \cdot\ 8^3\ +\ 9^{x-3}\ \cdot\ 6^3=9^x\ \frac{8^3}{9^3}\ +\ 9^x\ \frac{6^3}{9^3}=9^x\ \frac{728}{729}\ <\ 9^x \)
Deci \( 9^x\ >\ 8^x\ +\ 6^x, \ x\ge 3 \) si cum \( 4^x\ +\ 2^x\ >\ 1 \ \Longrightarrow\ RHS\ >\ LHS\ ,\ \ x\ge 3 \) . In concluzie \( x\in\{\ 0\ ,\ 1\ ,\ 2\ \} \) .