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

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Answer 1 · 2015-09-02T16:10:15

$x = sqrt(11)$

$log a + log b = log (ab)$
$log_a b = c <=> a^c = b$

$log (x+1) + log(x-1) = 1$
$log (x+1)(x-1) = 1$
Assuming $log$ base 10:
$x^2-1 = 10$
$x = +- sqrt(11)$

Since $x > 1$ , $x = sqrt(11)$

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