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Beliebt Algebra >

log_{1/2}(x^2-1)>= log_{1/2}(3x)

Lösung

lo g_{\frac{1}{2}} (x^{2} - 1) \geq lo g_{\frac{1}{2}} (3 x)

Lösung

1 < x \leq \frac{13 + 3}{2}

+2

Intervall-Notation

(1, \frac{13 + 3}{2}]

Dezimale

1 < x \leq 3.30277 \dots

Schritte zur Lösung

lo g_{\frac{1}{2}} (x^{2} - 1) \geq lo g_{\frac{1}{2}} (3 x)

Wende die log Regel an

x^{2} - 1 \leq 3 x and x^{2} - 1 > 0

x^{2} - 1 \leq 3 x and x^{2} - 1 > 0 : 1 < x \leq \frac{13 + 3}{2}

1 < x \leq \frac{13 + 3}{2}

Graph

Beliebte Beispiele

3x+1<2*log_{2}(x+4)

3 x + 1 < 2 \cdot lo g_{2} (x + 4)

log_{10}(x-2)(3x^2-x-1)>0

lo g_{10} (x - 2) (3 x^{2} - x - 1) > 0

2-log_{10}(4x)>= 0

2 - lo g_{10} (4 x) \geq 0

log_{2}(x-1)>-1

lo g_{2} (x - 1) > - 1

log_{x-2}(3x^2-x-1)>0

lo g_{x - 2} (3 x^{2} - x - 1) > 0