Solve: `(log)_((2x+3))(6x^2+23+21)+(log)_((3x+7))(4x^2+12 x+9)=4`

Explanation

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)`
`+2/(log_((2x+3))(3x+7))=4`
Let `log_((2x+3))(3x+7)=t`.Then
` 1+t+2/t = 4`
` or t^(2) = 3t+2 = 0`
` or (t-1)(t-2) = 0`
`rArr t=1, t = 2`
(i) if t = 1, then
` log_((2x+3))(3x+7)=1`
` or 3x+7 = 2x + 3`
Hence, x =- 4, which is not possible as ` 2x+3 gt 0 and 3x + 7 gt 0`.
(ii) if t = 2, then
` log_((2x+3))(3x+7)= 2`
` or 3x+7 = (2x+3)^(2)`
` or 4x^(2) + 9x + 2 = 0`
` or (x+2) (4x+1) = 0`
` rArr x =- 2 and x =- 1/4`
Sincex =- 2 is not possible, there is only one solution, ` x =- 1//4`.