JEE Main202120 Jul 2021Evening ShiftMathematicsBasics of MathematicsActual
The number of solutions of the equation log ( x + 1 ) 2 x 2 + 7 x + 5 + log ( 2 x + 5 ) ( x + 1 ) 2 - 4 = 0 , x > 0 , is
Correct answer
1
Step-by-step solution
We have, log ( x + 1 ) 2 x 2 + 7 x + 5 + log ( 2 x + 5 ) ( x + 1 ) 2 - 4 = 0 ,   x > 0 ⇒ log x + 1 2 x + 5 x + 1 + 2 log 2 x + 1 x + 1 = 4 ⇒ log x + 1 2 x + 5 + log x + 1 x + 1 + 2 log 2 x + 1 x + 1 = 4 ⇒ log x + 1 2 x + 5 + 1 + 2 log 2 x + 1 x + 1 = 4 ⇒ log x + 1 2 x + 5 + 2 log 2 x + 1 x + 1 = 3 ⇒ log x + 1 2 x + 5 + 2 log x + 1 2 x + 5 = 3 Put log ( x + 1 ) ( 2 x + 5 ) = t , then t + 2 t = 3 ⇒ t 2 - 3 t + 2 = 0 ⇒ t = 1 ,   2 Then, log ( x + 1 ) ( 2 x + 5 )