Find the value of $x$ which satisfies the equation: $log (x + 5) - log (3 x + 25) = log (x - 15) - log 17$ .

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Solution

$log (x + 5) - log (3 x + 25) = log (x - 15) - log 17$ is valid when
$x > - 5, x > - \frac{25}{3}, x > 15$
$\Rightarrow x > 15$
Rewrite given equation as
$log (\frac{x + 5}{3 x + 25}) = log (\frac{x - 15}{17}) [∵ log a - log b = log \frac{a}{b}]$
$\Rightarrow \frac{x + 5}{3 x + 25} = \frac{x - 15}{17}$
$\Rightarrow 17 x + 85 = 3 x^{2} - 20 x - 375$
$\Rightarrow 3 x^{2} - 37 x - 460 = 0$
$\Rightarrow x = 20$ as $x > 15$
Ans: 20

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