# Solve for x the following equation :log (2x + 3) (6x2 + 23x + 21) = 4 – log(3x + 7) (4x2 + 12x + 9)

10 years ago
Hello Student,
The given equation is
log (2x + 3) (6x2 + 23x + 21)
= 4 - log 3x + 7 ( 4x2 + 12x + 9)
⇒ log (2x + 3) (6x2 + 23x 21)
+ log (3x + 7) (4x2 + 12x + 9) = 4
⇒ log (2x + 3) (2x + 3) (3x + 7) + log (3x + 7) (2x + 3)2 x = 4
⇒ 1 + log (2x + 3) (3x + 7) + 2 log (3x + 7) (2x + 3) = 4
[ Using log ab = a + log b and log an = n log a]
NOTE THIS STEP
⇒ log (2x + 3) (3x + 7) + 2/ log (2x + 3) (3x + 7) = 3 [ Using logab = 1/logba]
Let log log (2x + 3) (3x + 7) = y
⇒ y + 2/y = 3 ⇒ y2 – 3y + 2 = 0
⇒ (y – 1) (y – 2) = 0 ⇒ y = 1, 2
Substituting the values of y in (1), we get
⇒ log (2x + 3) (3x + 7) = 1 and log (2x + 3) (3x + 7) = 2
⇒ 3x + 7 = 2x + 3 and 3x + 7 = (2x + 3)2
⇒ x = -4 and 4x2 + 9x + 2 = 0
⇒ x = -4 and (x + 2) (4x + 1) = 0
⇒ x = -4 and x = 02, x = -1/4
As log ax is defined for x > 0 and a > 0 (a ≠ 1), the possible value of x should satisfy all of the following inequalities :
⇒ 2x + 3 > 0 and 3x + 7 > 0
Also 2x + 3 ≠ 1 and 3x + 7 ≠ 1
Out of x = -4, x = -2 and x = -1/4 only x = -1/4
Satisfies the above inequalities.
So only solutions is x = -1/4.

Thanks