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

If log⁡(−2x)=2log⁡(x+1)\log (-2x)=2\log (x+1), then xxcan be equal to
A. −2+3-2+\sqrt{3}
B. −4+23-4+2\sqrt{3}
C. −2−3-2-\sqrt{3}
D. −4−23-4-2\sqrt{3}

Explanation

Solution

- 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 xx. Here we use the property that mlog⁡n=log⁡nmm\log n=\log {{n}^{m}}, so we can write 2log⁡(x+1)=log⁡(x+1)22\log (x+1)=\log {{(x+1)}^{2}}. And another property of logarithm that if log⁡ax=log⁡ay⇒x=y{{\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)=2log⁡(x+1)\log (-2x)=2\log (x+1)
Now using the property
mlog⁡n=log⁡nmm\log n=\log {{n}^{m}}
we can write the right-hand side term as
2log⁡(x+1)=log⁡(x+1)22\log (x+1)=\log {{(x+1)}^{2}}
So, we have
log⁡(−2x)=log⁡(x+1)2−−−−(1)\log (-2x)=\log {{(x+1)}^{2}}----(1)
Here we can assume that the base of logarithm is 10. Now using the property
log⁡ax=log⁡ay⇒x=y{{\log }_{a}}x={{\log }_{a}}y\Rightarrow x=y
We can write
−2x=(x+1)2−−−−−(2)-2x={{(x+1)}^{2}}-----(2)
Now as we know that
(a+b)2=a2+b2+2ab{{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab
Hence, we can write the equation (2)(2) as
−2x=x2+12+2x ⇒x2+4x+1=0 \begin{aligned} & -2x={{x}^{2}}+{{1}^{2}}+2x \\\ & \Rightarrow {{x}^{2}}+4x+1=0 \\\ \end{aligned}
Using the formula of quadratic equation,
if ax2+bx+c=0,a{{x}^{2}}+bx+c=0, then its roots is given by
x1,2=−b±b2−4(a)(c)2(a){{x}_{1,2}}=\dfrac{-b\pm \sqrt{{{b}^{2}}-4(a)(c)}}{2(a)}
Hence, we can write
x1,2=−4±42−4(1)(1)2(1) x1,2=−4±16−42(1) x1,2=−4±122(1) x1,2=−4±232(1) \begin{aligned} & {{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{aligned}
So, we get two values of xx, when we take positive sign, we have
x1=−2+3{{x}_{1}}=-2+\sqrt{3}
When we take negative sign, we get the value of xxas
x2=−2−3{{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 xxin each case is negative. As xxis negative, −x-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 ax=N{{a}^{x}}=N,xx is called logarithm of NNto the base aa
If ax=N then x=log⁡aN{{a}^{x}}=N\text{ then }x={{\log }_{a}}N
Here N is a positive number and aais a positive number other than 1.