Sol. The given equation can be written as

log (2x+3) (2x + 3) (3x + 7) + log (3x+7) (2x+3) 2 = 4

or log (2x+3) (2x + 3) + log (2x + 3) (3x + 7) +

= 4

Let log (2x+3) (3x + 7) = t

∴ 1 + t + = 4

or t 2 –3t + 2 = 0

or (t –1) (t – 2) = 0

∴ t = 1, t = 2

if t = 1 then log (2x+3) (3x + 7) = 1

∴ 3x + 7 = 2x + 3

∴ x = – 4 But 2x + 3 > 0 & 3x + 7 > 0

∴ x > – 3/2 & x > – 7/3

Hence no solution for t = 1

if t = 2 log (2x+3) (3x + 7) = 2

then or 3x + 7 = (2x + 3) 2

or 4x 2 + 9x + 2 = 0

∴ (x + 2) (4x + 1) = 0

∴ x = –2 and x = – (  x < – 3/2)

∴ only one solution x = –