The number of roots of the equation log(-2x) | vedclass

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Question

(C) Given equation: $\log(-2x) = 2\log(x+1)$.For the equation to be defined,the arguments of the logarithm must be positive:$-2x > 0 \Rightarrow x < 0

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) Given equation: $\log(-2x) = 2\log(x+1)$.For the equation to be defined,the arguments of the logarithm must be positive:$-2x > 0 \Rightarrow x  0 \Rightarrow x > -1$.Thus,the domain is $x \in (-1, 0)$.Applying logarithm properties: $\log(-2x) = \log((x+1)^2)$.Equating the arguments: $-2x = (x+1)^2$.$-2x = x^2 + 2x + 1 \Rightarrow x^2 + 4x + 1 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$x = \frac{-4 \pm \sqrt{16 - 4}}{2} = \frac{-4 \pm \sqrt{12}}{2} = -2 \pm \sqrt{3}$.Check the values against the domain $x \in (-1, 0)$:$x_1 = -2 + \sqrt{3} \approx -2 + 1.732 = -0.268$ (This is in the domain).$x_2 = -2 - \sqrt{3} \approx -2 - 1.732 = -3.732$ (This is not in the domain).Therefore,there is only $1$ valid root.

The number of roots of the equation $\log(-2x) = 2\log(x+1)$ is:

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