What is log_{1/3}(2x)<2-log_{1/3}((1-x)/x) ?

The solution to log_{1/3}(2x)<2-log_{1/3}((1-x)/x) is 0<x< 17/18

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

log_{1/3}(2x)<2-log_{1/3}((1-x)/x)

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

0 < x < \frac{17}{18}

Interval Notation

(0, \frac{17}{18})

Decimal Notation

0 < x < 0.94444 \dots

Solution steps

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

Move

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

to the left side

lo g_{\frac{1}{3}} (2 x) + lo g_{\frac{1}{3}} (\frac{1 - x}{x}) < 2

Apply log rules

2 (1 - x) > \frac{1}{9}

2 (1 - x) > \frac{1}{9} : x < \frac{17}{18}

x < \frac{17}{18}

Find positive values for logs:

0 < x < 1

Combine the intervals

x < \frac{17}{18} and 0 < x < 1

Merge Overlapping Intervals

0 < x < \frac{17}{18}

7 z^{2} + 8 z + 3 = 3 z^{2}

(- 1)^{\frac{1}{4}}

0 - (- 4)

\frac{1}{3} \cdot 5

y^{2} - 13 y + 36 = 0

What is log_{1/3}(2x)<2-log_{1/3}((1-x)/x) ?
The solution to log_{1/3}(2x)<2-log_{1/3}((1-x)/x) is 0<x< 17/18