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Grade:12

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Satyajit Samal
askIITians Faculty 34 Points
9 years ago
The given equation can be written in the form:
\log ^{ 2 }{ \left( \frac { x+4 }{ x } \right) }+\log ^{ 2 }{ \left( \frac { x }{ x+4 } \right) }=2\log ^{ 2 }\left( { \frac { 3-x }{ x-1 } } \right) \\ \Rightarrow 2\log ^{ 2 }{ \left( \frac { x+4 }{ x } \right) }=2\log ^{ 2 }\left( { \frac { 3-x }{ x-1 } } \right)
Now\log xis defined forx > 0. So,\frac{x+4}{x}> 0 \quad and \quad \frac{3-x}{x-1} > 0
Solving these two we get1 < x < 3…......................(1)
To solve\log ^{ 2 }{ \left( \frac { x+4 }{ x } \right) }=\log ^{ 2 }\left( { \frac { 3-x }{ x-1 } } \right), assume left hand side as t^{2}and right hand side asm^{2}. Then we get\left ( t-m \right )\left ( t+m \right )=0 \\ \Rightarrow t=m\; or\; t=-m.
\log { \left( \frac { x+4 }{ x } \right) }=\log \left( { \frac { 3-x }{ x-1 } } \right) \\ \Rightarrow \log \left ( \frac{x+4}{x}.\frac{x-1}{3-x} \right )=0 \\ \Rightarrow x = \pm \sqrt{2}….................(2)
From (1) and (2)x= \sqrt{2}is a possible solution.
Similarly solving\log { \left( \frac { x+4 }{ x } \right) }=-\log \left( { \frac { 3-x }{ x-1 } } \right), we getx= \pm \sqrt {6}..........(3)
From (1) and (3)x= \sqrt {6}is a possible solution.
Hence we get two solutions to the given equation:\sqrt{2}\;, \sqrt {6}.

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