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

The solution to log_{1/2}(x^2-1)>= log_{1/2}(3x) is 1<x<= (sqrt(13)+3)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Interval Notation

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

Decimal Notation

1 < x \leq 3.30277 \dots

Solution steps

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

Apply log rules

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}

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

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

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

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

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

What is log_{1/2}(x^2-1)>= log_{1/2}(3x) ?
The solution to log_{1/2}(x^2-1)>= log_{1/2}(3x) is 1<x<= (sqrt(13)+3)/2