$l o g (1 + x) - l o g (1 - x) = 1$
Find the value of x.

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Solution

$l o g (1 + x) - l o g (1 - x) = 1$

$l o g (\frac{1 + x}{1 - x}) = 1$

$\frac{1 + x}{1 - x} = 10^{1}$

(on taking Anti log)

$1 + x = 10 - 10 x$

$11 x = 9$

$x = \frac{9}{11}$ .

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