Question

The number of roots of the equation \log (-2x)
=2\log (x+1)
are [AMU 2001]
A) 3 B) 2 C) 1 D) None of these

Answer 1

option "B" is the correct answer


so 2 root of the equation..

I think my answer is capable to clear your confusion..

Answer 2

The given logarithm expression:

$\log (-2x)$ = 2$\log (x+1)$

We have to find, the number of roots of the given expression.

Solution:

∴ $\log (-2x)$ = 2$\log (x+1)$

Using the logarithm identity:

$n\log m=\log m^n$

⇒ $\log (-2x)$ = $\log (x+1)^2$

⇒ - 2x = $(x+1)^2$

⇒ - 2x = $x^2$ + 2x + 1 [∵ $(a+b)^{2} =a^{2} +2ab+b^{2}$]

⇒ $x^2$ + 4x + 1 = 0

Here, a = 1, b = 4 and c = 1

∴ Discriminant, D = $b^{2} -4ac$

= $4^{2}$ - 4(1)(1)

= 16 - 4

= 12 > 0, the two roots are real and unequal.

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

Thus, the required "option B) 2" is correct.