Question

Ecuația din imagine, vă rog !.

Answer 1

Explicație pas cu pas:

$log(5) 0.2 = - 1 \\ proba \\ 5 ^{ - 1} = \frac{1}{5} = 0.2 \\ log(3) (3 {}^{x {}^{2} - 13x + 28} + \frac{2}{9} ) = - 1 \\ 3 {}^{ - 1} = 3 {}^{x {}^{2} - 13x + 28} + \frac{2}{9} \\ 3 {}^{ - 1} = \frac{ 9(3 {}^{x {}^{2} - 13x + 28}) + 2}{9} \\ 3 {}^{ - 1} \times 3 {}^{2} = 9(3 {}^{x {}^{2} - 13x + 28}) + 2\\ 3 = 9(3 {}^{x {}^{2} - 13x + 28}) + 2 \\ 9(3 {}^{x {}^{2} - 13x + 28}) = 1\\ 3 {}^{x {}^{2} - 13x + 28} = \frac{1}{9} \\ 3 {}^{x {}^{2} - 13x + 28} = 3 {}^{ - 2} \\ x {}^{2} - 13x + 28 = - 2 \\ x {}^{2} - 13x + 30 = 0 \\ (x - 10)(x - 3) = 0 \\ x1 = 10 \\ x2 = 3$