Ecuatie in numere reale
Posted: Tue Jan 19, 2010 10:41 pm
Sa se rezolve ecuatia:
\( 3^{x+1}=(x-1)(x-3) \)
\( 3^{x+1}=(x-1)(x-3) \)
1) Daca \( x\in (-\infty,1) \), \( x=0 \) solutie unica.
2) Daca \( x\in(3,+\infty) \) nu avem solutii, deoarece daca \( x\in[n,n+1),n\ge 3 \), atunci \( 3^{x+1}\ge3^{n+1}>n(n-2)>(x-1)(x-3) \).