Question

If $\log (-2x)=2\log (x+1)$, then $x$can be equal to
A. $-2+\sqrt{3}$
B. $-4+2\sqrt{3}$
C. $-2-\sqrt{3}$
D. $-4-2\sqrt{3}$

Answer 1

- Hint: In such a type of question there are two functions one is logarithmic and other is algebraic. So first of all, we have to use the properties of logarithms and then we use properties of algebra to find the value of $x$. Here we use the property that $m\log n=\log {{n}^{m}}$, so we can write $2\log (x+1)=\log {{(x+1)}^{2}}$. And another property of logarithm that if ${{\log }_{a}}x={{\log }_{a}}y\Rightarrow x=y$.
Once we equate the two terms, we have to solve the equation in order to get the solution of the given question.

Complete step-by-step solution -
It is given from question
$\log (-2x)=2\log (x+1)$
Now using the property
 $m\log n=\log {{n}^{m}}$
we can write the right-hand side term as
$2\log (x+1)=\log {{(x+1)}^{2}}$
So, we have
$\log (-2x)=\log {{(x+1)}^{2}}----(1)$
Here we can assume that the base of logarithm is 10. Now using the property
${{\log }_{a}}x={{\log }_{a}}y\Rightarrow x=y$
We can write
$-2x={{(x+1)}^{2}}-----(2)$
Now as we know that
${{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$
Hence, we can write the equation $(2)$ as
$\begin{align}
  & -2x={{x}^{2}}+{{1}^{2}}+2x \\
 & \Rightarrow {{x}^{2}}+4x+1=0 \\
\end{align}$
Using the formula of quadratic equation,
if $a{{x}^{2}}+bx+c=0,$ then its roots is given by
${{x}_{1,2}}=\dfrac{-b\pm \sqrt{{{b}^{2}}-4(a)(c)}}{2(a)}$
Hence, we can write
$\begin{align}
  & {{x}_{1,2}}=\dfrac{-4\pm \sqrt{{{4}^{2}}-4(1)(1)}}{2(1)} \\
 & {{x}_{1,2}}=\dfrac{-4\pm \sqrt{16-4}}{2(1)} \\
 & {{x}_{1,2}}=\dfrac{-4\pm \sqrt{12}}{2(1)} \\
 & {{x}_{1,2}}=\dfrac{-4\pm 2\sqrt{3}}{2(1)} \\
\end{align}$
So, we get two values of $x$, when we take positive sign, we have
${{x}_{1}}=-2+\sqrt{3}$
When we take negative sign, we get the value of $x$as
${{x}_{2}}=-2-\sqrt{3}$
So, we get two solutions to the given equation.
Hence option A and C both are correct

Note: It should be noted that values of $x$in each case is negative. As $x$is negative, $-x$ is positive , which is in accordance with the definition of logarithm, as logarithm of negative number is not defined.
The logarithm of any number to a given base is the index of the power to which the base must be raised in order to equal to the given number. If ${{a}^{x}}=N$,$x$ is called logarithm of $N$to the base $a$
If ${{a}^{x}}=N\text{ then }x={{\log }_{a}}N$
Here N is a positive number and $a$is a positive number other than 1.