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Grade:12

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Satyajit Samal
askIITians Faculty 34 Points
9 years ago
Using logarithm property the given equation can be written as:\log_{4} \left ( x+2 \right )^{3} +3 =\log _{4}\left \{ \left ( 4-x \right )\left ( 6+x \right ) \right \}^{3} \\ \Rightarrow 3 \log _{4}\frac{\left ( 4-x \right )\left ( 6+x \right )}{x+2}=3 \\ \Rightarrow \left ( 4-x \right )\left ( 6+x \right )=4\left ( x+2 \right ) \\ \Rightarrow x^2 +6x -16 = 0 \\ \Rightarrow x = -8 \quad or \quad x=2
Also using the domain of log x which is define only when x > 0 , we get from three log functions that x > -2 , x <4 and x > -6 .
Hence only x =2 is a valid solution.

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