Question

Sa se rezolve ecuatia:
$log_{x + 1}(2x {}^{3} + 2x {}^{2} - 3x + 1 ) = 3$
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Answer 1

Răspuns:

Explicație pas cu pas:

log(x+1 ;2x^3 +2x^2 -3x +1) = log(x+1; (x+1)^3)

2x^3 +2x^2 -3x +1 = (x+1)^3 = (x+1)(x^2+2x +1)

2x^3 +2x^2 -3x +1 = x^3 +2x^2 +x +x^2+2x +1

2x^3 +2x^2 -3x +1 = x^3 +3x^2 +3x +1

x^3 -x^2 -6x = 0

x(x^2 -x -6) = 0

x1 = 0, nu convie,  x+1 = 0+1 = 1, baza sa fie ≠ 1

x^2 -x -6 = 0,  delta = 1 +24 = 25

x2,3 = (1 -+5)/2

x2 = -4/2 = -2,  nu e  buna, val.de sub log e < 0

x3 = 6/2 = 3,  convine