Question

The set of real roots of the equation

log_{(5x + 4)} (2x + 3)³ – log_{(2x + 3)} (10x² + 23x + 12) = 1 is –

• A. {– 1}
• B. {– 3/5}
• C. Empty set
• D. {– 1/3}

Answer 1

Answer: (D)

{– 1/3}

Answer 2

We have

log_{(5x + 4)} (2x + 3)³ – log_{(2x + 3)}

(10x² + 23x + 12) = 1

⇒ 3 log_{(5x + 4)} (2x + 3) – log_{(2x + 3)} (5x + 4)

– log_{2x + 3} (2x + 3) = 1

⇒ 3 log_{(5x + 4)} (2x + 3) – log_{(2x + 3)} (5x + 4) = 2 ….(1)

⇒ 3y – $\frac{1}{y}$– 2 = 0 where y = log_{(5x + 4)} (2x + 3)

⇒ 3y² – 2y – 1 = 0 ⇒ y = – $\frac{1}{3}$, 1.

(1) ⇒ 2x + 3 > 0, 5x + 4 > 0, 5x + 4 ≠ 1, 2x + 3 ≠ 1

⇒ x > – 3/2, x > – 4/5, x ≠ – 3/5, x ≠ – 1

⇒ x > – 3/5.

Now 2x + 3 > 2$\left( - \frac{3}{5} \right)$+ 3 = $\frac{9}{5}$> 1

and 5x + 4 > 5$\left( - \frac{3}{5} \right)$+ 4 = 1

∴ log (2x + 3) > 0, log (5x + 4) > 0

y = log_{(5x + 4)} (2x + 3)

= $\frac{\log(2x + 3)}{\log(5x + 4)}$= $\frac{( + )}{( + )}$> 0

∴ y ≠ – $\frac{1}{3}$

∴ y = 1 ⇒ $\frac{\log(2x + 3)}{\log(5x + 4)}$= 1 ⇒ 2x + 3 = 5x + 4

⇒ x = – $\frac{1}{3}$.