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JEE Advanced
Exam
JEE Advanced
Subject
Mathematics
Chapter
Basics of Mathematics

JEE Advanced · Mathematics · Basics of Mathematics

The value(s) of x satisfying 1- _9(x+1)^2=(1)/(2) _ 3 ((x+5)/(x+3))...

The value(s)ofxof x satisfying 1−log⁡9(x+1)2=12log⁡31-\log _9(x+1)^2=\frac{1}{2} \log _{\sqrt{3}}\left(\frac{x+5}{x+3}\right) is/are

Options

  1. A1
  2. B-2
  3. C-7
  4. D-4

Correct answer

C. -7

Step-by-step solution

1−log⁡32(x+1)2=12log⁡31/2(x+5x+3)⇒1−22log⁡3∣x+1∣=22log⁡3(x+5x+3)⇒log⁡3(3∣x+1∣)=log⁡3(x+5x+3)⇒3∣x+1∣=x+5x+3 Case- I x+1>0⇒x>−1⇒3(x+3)=(x+1)(x+5)⇒x2+3x−4=0⇒x=−4 or x=1x=−4 rejected (∵x>−1)∴x=1 Case - II x+1<0⇒x<−1⇒3(x+3)=−(x+1)(x+5)⇒x2+9x+14=0⇒x=−2 or x=−7∴ Set of value of x={−7,−2,1}\begin{aligned} & 1-\log _{3^2}(x+1)^2=\frac{1}{2} \log _{3^{1 / 2}}\left(\frac{x+5}{x+3}\right) \\ & \Rightarrow 1-\frac{2}{2} \log _3|x+1|=\frac{2}{2} \log _3\left(\frac{x+5}{x+3}\right) \\ & \Rightarrow \log _3\left(\frac{3}{|x+1|}\right)=\log _3\left(\frac{x+5}{x+3}\right) \\ & \Rightarrow \frac{3}{|x+1|}=\frac{x+5}{x+3} \\ & \text { Case- I } \qquad x+1\gt0 \Rightarrow x\gt-1 \\ & \Rightarrow 3(x+3)=(x+1)(x+5) \\ & \Rightarrow x^2+3 x-4=0 \\ & \Rightarrow x=-4 \text { or } x=1 \\ & x=-4 \text { rejected } \qquad (\because x\gt-1) \\ & \therefore \mathrm{x}=1 \\ & \text { Case - II } \qquad \mathrm{x}+1 \lt 0 \Rightarrow \mathrm{x} \lt -1 \\ & \Rightarrow 3(x+3)=-(x+1)(x+5) \\ & \Rightarrow \mathrm{x}^2+9 \mathrm{x}+14=0 \\ & \Rightarrow x=-2 \text { or } x=-7 \\ & \therefore \text { Set of value of } x=\{-7,-2,1\} \end{aligned}
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