The number of roots of the equation log -2 x -2 log x -1 areA- 3B- 2C- 1D- 0

The correct option is C 1 Given, log( 2x) = 2 log(x+1) xϵ( 1,0) ⇒ 2x = (x+1)^2 ⇒ x^2+4x+1=0 ⇒ x = 4±√(16 4)/2 ⇒ x = 4±√(12)/2 ⇒x = 2 ± √(3) ⇒ x = ( ...

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Standard XII

Mathematics

Location of Roots

The number of...

Question

The number of roots of the equation log(-2x) = 2 log(x+1) are

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Solution

The correct option is C
1

Given, log(-2x) = 2 log(x+1) $x ϵ (- 1, 0)$

⇒ -2x = $(x + 1)^{2}$ ⇒ $x^{2} + 4 x + 1 = 0$

⇒ $x = \frac{- 4 \pm \sqrt{16 - 4}}{2}$ ⇒ $x = \frac{- 4 \pm \sqrt{12}}{2}$

⇒ $x = - 2 \pm \sqrt{3}$ ⇒ $x = (- 2 + \sqrt{3}), (- 2 - \sqrt{3})$

only one solution

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The number of roots of the equation log(-2x) = 2 log(x+1) are

The correct option is C 1 Given, log(-2x) = 2 log(x+1) xϵ(−1,0) ⇒ -2x = (x+1)2 ⇒ x2+4x+1=0 ⇒ x=−4±√16−42 ⇒ x=−4±√122 ⇒x=−2±√3 ⇒ x=(−2+√3),(−2−√3) only one solution

The correct option is C
1

Given, log(-2x) = 2 log(x+1) $x ϵ (- 1, 0)$

⇒ -2x = $(x + 1)^{2}$ ⇒ $x^{2} + 4 x + 1 = 0$

⇒ $x = \frac{- 4 \pm \sqrt{16 - 4}}{2}$ ⇒ $x = \frac{- 4 \pm \sqrt{12}}{2}$

⇒ $x = - 2 \pm \sqrt{3}$ ⇒ $x = (- 2 + \sqrt{3}), (- 2 - \sqrt{3})$

only one solution