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Question Number 109721 by Algoritm last updated on 25/Aug/20

	Answered by 1549442205PVT last updated on 25/Aug/20
	$We need have the condition that 6−x>0 for logarithm is meaning.Then log_3 (6−x)(√(x^2 +2x+6))=x ⇔log_3 (6−x)(√(x^2 +2x+6)) −x=0 Put f(x)=(6−x)(√(x^2 +2x+6)) −x f ′(x)=(1/((ln3)(6−x)(√(x^2 +2x+6)) ))×[(6−x)(√(x^2 +2x+6)) ]′−1 ={(1/(ln3(6−x)(√(x^2 +2x+6)) ))×[(6−x).((x+1)/()(√(x^2 +2x+6)) ))−(√(x^2 +2x+6)) ]}−1 ={(1/(ln3(6−x)(√(x^2 +2x+6)) ))×((−x^2 +5x+6−(x^2 +2x+6))/( (√(x^2 +2x+6))))}−1 ={(1/(ln3(6−x)(√(x^2 +2x+6)) ))×((−2x^2 +3x)/( (√(x^2 +2x+6))))}−1 =((−2x^2 +3x)/(ln3(6−x)(x^2 +2x+6)))−1 we prove that f ′(x)<0 ∀x<6.Indeed, f ′(x)<0⇔((−2x^2 +3x)/(ln3(6−x)(x^2 +2x+6)))−1<0 ⇔ln3(6−x)(x^2 +2x+6)>−2x^2 +3x Since ln3>1,it is enough to prove that (6−x)(x^2 +2x+6)>−2x^2 +3x (∗) ⇔−x^3 +4x^2 +6x+36>−2x^2 +3x ⇔−x^3 +6x+3x+36>0 Put h(x)=−x^3 +6x^2 +3x+36 h′(x)=−3x^2 +12x+3=0 ⇔−x^2 +4x+1=0 Δ′=4+1=5 x_(1,2) =2±(√5) .We have variable tablet determinant (((−∞),,(2−(√5)),,,(2+(√5)),,,6),((h′(x)),(−↘),,,+,,↘,−,),((h(x)),,(35.64),↗,,(80.36),,,(54))) From above tablet we have :h(6)=54 h_(min) (x)=35.6,h_(max) (x)=80.3,x∈(−∞;6) .This shows that h(x)>0∀x<6 Hence the inequality(∗) is true ∀x<6 which means f ′(x)<0∀x<6 ,so we infer f(x)is decreasing function that means the equation f(x)=0 has no more one roots.Solve the f(x)=0 ⇔log_3 (6−x)(√(x^2 +2x+6)) −x=0 we get the root x=2.440710525 Thus the give eqution has unique root x=2.440710525$
	$We need have the condition that$ $6 - x > 0 for logarithm is meaning . Then$ $\log_{3} (6 - x) \sqrt{x^{2} + 2 x + 6} = x$ $\Leftrightarrow \log_{3} (6 - x) \sqrt{x^{2} + 2 x + 6} - x = 0$ $Put f (x) = (6 - x) \sqrt{x^{2} + 2 x + 6} - x$ $f^{'} (x) = \frac{1}{(\ln 3) (6 - x) \sqrt{x^{2} + 2 x + 6}} \times {[(6 - x) \sqrt{x^{2} + 2 x + 6}]}^{'} - 1$ $= {\frac{1}{\ln 3 (6 - x) \sqrt{x^{2} + 2 x + 6}} \times [(6 - x) . \frac{x + 1}{) \sqrt{x^{2} + 2 x + 6}} - \sqrt{x^{2} + 2 x + 6}]} - 1$ $= {\frac{1}{\ln 3 (6 - x) \sqrt{x^{2} + 2 x + 6}} \times \frac{- x^{2} + 5 x + 6 - (x^{2} + 2 x + 6)}{\sqrt{x^{2} + 2 x + 6}}} - 1$ $= {\frac{1}{\ln 3 (6 - x) \sqrt{x^{2} + 2 x + 6}} \times \frac{- {2 x}^{2} + 3 x}{\sqrt{x^{2} + 2 x + 6}}} - 1$ $= \frac{- {2 x}^{2} + 3 x}{\ln 3 (6 - x) (x^{2} + 2 x + 6)} - 1$ $we prove that f^{'} (x) < 0 \forall x < 6 . Indeed,$ $f^{'} (x) < 0 \Leftrightarrow \frac{- {2 x}^{2} + 3 x}{\ln 3 (6 - x) (x^{2} + 2 x + 6)} - 1 < 0$ $\Leftrightarrow \ln 3 (6 - x) (x^{2} + 2 x + 6) > - {2 x}^{2} + 3 x$ $Since \ln 3 > 1, it is enough to prove that$ $(6 - x) (x^{2} + 2 x + 6) > - {2 x}^{2} + 3 x ()$ $\Leftrightarrow - x^{3} + {4 x}^{2} + 6 x + 36 > - {2 x}^{2} + 3 x$ $\Leftrightarrow - x^{3} + 6 x + 3 x + 36 > 0$ $Put h (x) = - x^{3} + {6 x}^{2} + 3 x + 36$ $h^{'} (x) = - {3 x}^{2} + 12 x + 3 = 0$ $\Leftrightarrow - x^{2} + 4 x + 1 = 0 Δ^{'} = 4 + 1 = 5$ $x_{1, 2} = 2 \pm \sqrt{5} . We have variable tablet$ $\| \begin{matrix} - \infty & 2 - \sqrt{5} & 2 + \sqrt{5} & 6 \\ h^{'} (x) & - ↘ & + & ↘ & - \\ h (x) & 35.64 & ↗ & 80.36 & 54 \end{matrix} \|$ $From above tablet we have : h (6) = 54$ $h_{\min} (x) = 35.6, h_{\max} (x) = 80.3, x \in (- \infty; 6)$ $. This shows that h (x) > 0 \forall x < 6$ $Hence the inequality () is true \forall x < 6$ $which means f^{'} (x) < 0 \forall x < 6, so we infer$ $f (x) is decreasing function that means$ $the equation f (x) = 0 has no more$ $one roots . Solve the f (x) = 0$ $\Leftrightarrow \log_{3} (6 - x) \sqrt{x^{2} + 2 x + 6} - x = 0$ $we get the root x = 2.440710525$ $Thus the give eqution has unique$ $root x = 2.440710525$
	Commented by Algoritm last updated on 25/Aug/20

	$x \approx 3.8$
	Commented by 1549442205PVT last updated on 25/Aug/20

	$You is false, check by calculator again!$
	Commented by Algoritm last updated on 25/Aug/20

	$sir x = 2.44 no correct$
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