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

       @bemath@  log _(∣2x−(1/2)∣) (x+1+(1/x))≥log _(∣2x−(1/2)∣) (x^2 +1+(1/x^2 ))

$$\:\:\:\:\:\:\:@{bemath}@ \\ $$$$\mathrm{log}\:_{\mid\mathrm{2}{x}−\frac{\mathrm{1}}{\mathrm{2}}\mid} \left({x}+\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\geqslant\mathrm{log}\:_{\mid\mathrm{2}{x}−\frac{\mathrm{1}}{\mathrm{2}}\mid} \left({x}^{\mathrm{2}} +\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right) \\ $$

Commented by bemath last updated on 10/Aug/20

thank you

$${thank}\:{you} \\ $$

Answered by bobhans last updated on 10/Aug/20

      ∔bobhans∔  log _(∣2x−(1/2)∣) (x+1+(1/x)) ≥ log _(∣2x−(1/2)∣) (x^2 +1+(1/x^2 ))   { ((x+1+(1/x)>0)),((x^2 +1+(1/x^2 )>0)),((∣2x−(1/2)∣>0)) :}   ⇒ { ((x>0)),((x≠0)),((x≠(1/4))) :}  ⇒((x^3 +x−x^4 −1)/((2x−(3/2))(2x+(1/2)))) ≥ 0 ...(×(−1)  ⇔ (((x+1+(1/x))−(x^2 +1+(1/x^2 )))/(∣2x−(1/2)∣−1)) ≥ 0...(×x^2 )  ⇒((x^3 +x−x^4 −1)/((2x−(3/2))(2x+(1/2)))) ≥ 0 ...(×(−1)  ⇒((x^4 −x^3 −x+1)/((2x−(3/2))(2x+(1/2)))) ≤ 0...(×(2x+(1/2)))  ⇒(((x−1)(x^3 −1))/((2x−(3/2)))) ≤ 0   ⇒(((x−1)^2 (x^2 +x+1))/((2x−(3/2)))) ≤0 ; x^2 +x+1 >0 ∀x∈R  ⇒(((x−1)^2 )/((2x−(3/2)))) ≤ 0 ⇒solution   x∈(0, (1/4)) ∪ ((1/4),(3/4)) ∪ {1}

$$\:\:\:\:\:\:\dotplus\mathrm{bobhans}\dotplus \\ $$$$\mathrm{log}\:_{\mid\mathrm{2x}−\frac{\mathrm{1}}{\mathrm{2}}\mid} \left(\mathrm{x}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}\right)\:\geqslant\:\mathrm{log}\:_{\mid\mathrm{2x}−\frac{\mathrm{1}}{\mathrm{2}}\mid} \left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right) \\ $$$$\begin{cases}{\mathrm{x}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}>\mathrm{0}}\\{\mathrm{x}^{\mathrm{2}} +\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }>\mathrm{0}}\\{\mid\mathrm{2x}−\frac{\mathrm{1}}{\mathrm{2}}\mid>\mathrm{0}}\end{cases}\:\:\:\Rightarrow\begin{cases}{\mathrm{x}>\mathrm{0}}\\{\mathrm{x}\neq\mathrm{0}}\\{\mathrm{x}\neq\frac{\mathrm{1}}{\mathrm{4}}}\end{cases} \\ $$$$\Rightarrow\frac{\mathrm{x}^{\mathrm{3}} +\mathrm{x}−\mathrm{x}^{\mathrm{4}} −\mathrm{1}}{\left(\mathrm{2x}−\frac{\mathrm{3}}{\mathrm{2}}\right)\left(\mathrm{2x}+\frac{\mathrm{1}}{\mathrm{2}}\right)}\:\geqslant\:\mathrm{0}\:...\left(×\left(−\mathrm{1}\right)\right. \\ $$$$\Leftrightarrow\:\frac{\left(\mathrm{x}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}\right)−\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)}{\mid\mathrm{2x}−\frac{\mathrm{1}}{\mathrm{2}}\mid−\mathrm{1}}\:\geqslant\:\mathrm{0}...\left(×\mathrm{x}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\frac{\mathrm{x}^{\mathrm{3}} +\mathrm{x}−\mathrm{x}^{\mathrm{4}} −\mathrm{1}}{\left(\mathrm{2x}−\frac{\mathrm{3}}{\mathrm{2}}\right)\left(\mathrm{2x}+\frac{\mathrm{1}}{\mathrm{2}}\right)}\:\geqslant\:\mathrm{0}\:...\left(×\left(−\mathrm{1}\right)\right. \\ $$$$\Rightarrow\frac{\mathrm{x}^{\mathrm{4}} −\mathrm{x}^{\mathrm{3}} −\mathrm{x}+\mathrm{1}}{\left(\mathrm{2x}−\frac{\mathrm{3}}{\mathrm{2}}\right)\left(\mathrm{2x}+\frac{\mathrm{1}}{\mathrm{2}}\right)}\:\leqslant\:\mathrm{0}...\left(×\left(\mathrm{2x}+\frac{\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$$$\Rightarrow\frac{\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{3}} −\mathrm{1}\right)}{\left(\mathrm{2x}−\frac{\mathrm{3}}{\mathrm{2}}\right)}\:\leqslant\:\mathrm{0}\: \\ $$$$\Rightarrow\frac{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)}{\left(\mathrm{2x}−\frac{\mathrm{3}}{\mathrm{2}}\right)}\:\leqslant\mathrm{0}\:;\:\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\:>\mathrm{0}\:\forall\mathrm{x}\in\mathbb{R} \\ $$$$\Rightarrow\frac{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} }{\left(\mathrm{2x}−\frac{\mathrm{3}}{\mathrm{2}}\right)}\:\leqslant\:\mathrm{0}\:\Rightarrow\mathrm{solution}\: \\ $$$$\mathrm{x}\in\left(\mathrm{0},\:\frac{\mathrm{1}}{\mathrm{4}}\right)\:\cup\:\left(\frac{\mathrm{1}}{\mathrm{4}},\frac{\mathrm{3}}{\mathrm{4}}\right)\:\cup\:\left\{\mathrm{1}\right\}\: \\ $$$$ \\ $$

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