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

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

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Answer 1 · 2015-09-02T16:10:15

$x = \sqrt{11}$ $x = sqrt(11)$

Explanation:

$log a + log b = log (a b)$ $log a + log b = log (ab)$
${log}_{a} b = c \Leftrightarrow a^{c} = b$ $log_a b = c <=> a^c = b$

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

Since $x > 1$ $x > 1$ , $x = \sqrt{11}$ $x = sqrt(11)$

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

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