The number of solutions of the equation $\log _{(x+1)}\left(2 x^{2}+7 x+5\right)+\log _{(2 x+5)}(x+1)^{2}-4=0, x>0$, is

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Basic Mathematics

The number of solutions of the equation

lo g_{(x + 1)} (2 x^{2} + 7 x + 5) + lo g_{(2 x + 5)} (x + 1)^{2} - 4 = 0, x > 0

, is

Answer: 1

Since

2 x^{2} + 7 x + 5 = (2 x + 5) (x + 1)

, put

t = lo g_{x + 1} (2 x + 5)

. Then

lo g_{x + 1} [(2 x + 5) (x + 1)] = t + 1, lo g_{2 x + 5} (x + 1)^{2} = \frac{2}{t}

and the equation becomes

t + 1 + \frac{2}{t} - 4 = 0 ⟹ t^{2} - 3 t + 2 = 0 ⟹ t = 1 or t = 2

For

t = 1

2 x + 5 = x + 1

gives

x = - 4

, which fails

x > 0

. For

t = 2

2 x + 5 = (x + 1)^{2}

gives

x^{2} = 4

, and only

x = 2

satisfies

x > 0

; there the bases

x + 1 = 3

and

2 x + 5 = 9

are both admissible. Hence the equation has exactly one solution.

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Integer answer