log_{2}(2+x)<1

Lösung

lo g_{2} (2 + x) < 1

- 2 < x < 0

Intervall-Notation

(- 2, 0)

Schritte zur Lösung

lo g_{2} (2 + x) < 1

Wende die log Regel an

2 + x < 2 and 2 + x > 0

2 + x < 2 and 2 + x > 0 : - 2 < x < 0

- 2 < x < 0

x + 2 \geq x

\frac{3}{( 2 - x )} - 2 - x < 2

x - 5 - 9 - x > 1

x^{2} < x + 1

x > 1 - x