How do you solve $log(x+1) - log(x-1)=1$ ?

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Answer 1 · 2015-09-02T16:13:11

$x = 11/9$

$log a - log b = log(a/b)$
$log_a b = c <=> a^c = b$

$log (x+1) - log (x-1) = 1$
$log (x+1)/(x-1) = 1$
Assuming $log$ base 10:
$(x+1)/(x-1) = 10$
$x+1 = 10(x-1)$
$x = 11/9$

How do you solve log(x+1) - log(x-1)=1log(x+1) - log(x-1)=1?