Question

If log 3 log (3x - 2) and log (3x +4) are in arithmetic
progression, then x is equal to

(a) 8/3
(b) log 3⁸
(c) log 2³
(d) 8​

Answer 1

Step-by-step explanation:

Given that, log(3), log(3x - 2), log(3x + 4) are in ap

so,

$log(3x - 2) - log(3) = log(3x + 4) - log(3x - 2)$

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

$= > (3x - 2)^{2} = 3(3x + 4)$

$= > 9 {x}^{2} - 12x + 4 = 9x + 12$

$= > 9 {x}^{2} - 21x - 8 = 0$

$= > 9 {x}^{2} - 24x + 3x - 8 = 0$

$= > 3x(3x - 8) + 1(3x - 8) = 0$

$= > (3x - 8)(3x + 1) = 0$

$= > x = \frac{8}{3} \: \: or \: \: x = - \frac{1}{3}$