Решите уравнение log2 (x^+3) = 1+log (x+3)

Log2 (x^2+3) = 1+log2 (x+3)

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ОДЗ:

{x^2+3>0; x e R

{x+3>0; x>-3

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log2 (x^2+3) = log2 (2) + log2 (x+3)

log2 (x^2+3) = log2 (2x+6)

x^2+3=2x+6

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

x^2-2x-3=0

D = (-2) ^2-4*1 * (-3) = 16

x1 = (2-4) / 2=-1

x2 = (2+4) / 2=3

Ответ: {-1; 3}