Question

Solve for x the following equation Log(6x^2 +23x+21) to the vse (2x+3) = 4 - log (4x^2+12x+9) to the base (3x+7). Explain in detail.

Answer 1

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 log_ab = 1/log_ba]

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.