Solve the following equations.
$(6 x - 5) | l n (2 x - 2.3) | = 8 l n (2 x + 2.3)$

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Solution

$(6 x - 5) | l n (2 x - 2.3) | = 8 l n (2 x + 2.3)$
We know that ln $x > 0 \forall x > 1$
and $l n x < 0 \forall_{0} < x < 1$
$∴ (6 x - 5) | l n (2 x + 2.3) | = 8 l n (2 x + 2.3)$
case I :
If $2 x + 2.3 > 1$
$\Rightarrow x \geq - 1.3 / 2$
$\Rightarrow x \geq - 13 / 20,$ then
$(6 x - 5) l n (2 x + 2.3) - 8 l n (2 x + 2.3) = 0$
$\Rightarrow l n (2 x + 2.3) [6 x - 13] = 0$
$∴ l n (2 x + 2.3) = 0$ or $6 x - 13 = 0$
$\Rightarrow 2 x + 2.3 = 1$ or $x = 13 / 6$
$x = - 13 / 20$ or $x = 13 / 6$
case 2 :
if $2 x + 2.3 < 1 \Rightarrow x < - 13 / 20$ then
$- (6 x - 5) l n (2 x + 2.3) - 8 l n (2 x + 2.3) = 0$
$\Rightarrow l n (2 x + 2.3) (6 x + 3) = 0$
$∴ l n (2 x + 2.3) = 0$ = or $6 x + 3 = 0$
$2 x + 2.3 = 1$ or $x = - 1 / 2$
$x = - 13 / 20$
However in the case 2 the allowed region is $a < - 13 / 20$
Hence $x = - 13 / 20$ & $- 1 / 2$ are not permissible
$∴ x = - 13 / 20$ & $x = 13 / 6$ are the only solution possible

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