bemath-log-2x-1-2-x-1-1-x-log-2x-1-2-x-2-1-1-x-2-

Question Number 106996 by bemath last updated on 10/Aug/20

@ b e m a t h @

\log_{∣ 2 x - \frac{1}{2} ∣} (x + 1 + \frac{1}{x}) ⩾ \log_{∣ 2 x - \frac{1}{2} ∣} (x^{2} + 1 + \frac{1}{x^{2}})

Commented by bemath last updated on 10/Aug/20

t h a n k y o u

Answered by bobhans last updated on 10/Aug/20

∔ bobhans ∔

\log_{∣ 2 x - \frac{1}{2} ∣} (x + 1 + \frac{1}{x}) ⩾ \log_{∣ 2 x - \frac{1}{2} ∣} (x^{2} + 1 + \frac{1}{x^{2}})

{\begin{cases} x + 1 + \frac{1}{x} > 0 \\ x^{2} + 1 + \frac{1}{x^{2}} > 0 \\ ∣ 2 x - \frac{1}{2} ∣> 0 \end{cases} \Rightarrow {\begin{cases} x > 0 \\ x \neq 0 \\ x \neq \frac{1}{4} \end{cases}

\Rightarrow \frac{x^{3} + x - x^{4} - 1}{(2 x - \frac{3}{2}) (2 x + \frac{1}{2})} ⩾ 0 \dots (\times (- 1)

\Leftrightarrow \frac{(x + 1 + \frac{1}{x}) - (x^{2} + 1 + \frac{1}{x^{2}})}{∣ 2 x - \frac{1}{2} ∣ - 1} ⩾ 0 \dots (\times x^{2})

\Rightarrow \frac{x^{3} + x - x^{4} - 1}{(2 x - \frac{3}{2}) (2 x + \frac{1}{2})} ⩾ 0 \dots (\times (- 1)

\Rightarrow \frac{x^{4} - x^{3} - x + 1}{(2 x - \frac{3}{2}) (2 x + \frac{1}{2})} ⩽ 0 \dots (\times (2 x + \frac{1}{2}))

\Rightarrow \frac{(x - 1) (x^{3} - 1)}{(2 x - \frac{3}{2})} ⩽ 0

\Rightarrow \frac{{(x - 1)}^{2} (x^{2} + x + 1)}{(2 x - \frac{3}{2})} ⩽ 0; x^{2} + x + 1 > 0 \forall x \in R

\Rightarrow \frac{{(x - 1)}^{2}}{(2 x - \frac{3}{2})} ⩽ 0 \Rightarrow solution

x \in (0, \frac{1}{4}) \cup (\frac{1}{4}, \frac{3}{4}) \cup {1}

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