InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Solve: `(log)_((2x+3))(6x^2+23+21)+(log)_((3x+7))(4x^2+12 x+9)=4` |
|
Answer» 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`. |
|