# How do you solve  log_10(x-1) - log_10(x+1) = 1?

Feb 11, 2016

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

Please see Comments by @Lotusbluete and @Tom below for conditions under which solution exists.
Mathematically stated: The solution exists only for the equation
${\log}_{10} | x - 1 | - {\log}_{10} | x + 1 | = 1$

#### Explanation:

On the RHS use the following Basic Log rule
log_b(m/n) = log_b(m) – log_b(n)
and on the LHS use the equality ${\log}_{10} 10 = 1$
${\log}_{10} \left(\frac{x - 1}{x + 1}\right) = {\log}_{10} 10$
Taking antilog of both sides

$\frac{x - 1}{x + 1} = 10$
$\implies \left(x - 1\right) = 10 \left(x + 1\right)$
$\implies x - 1 = 10 x + 10$
$\implies 9 x = - 11$
$\implies x = - \frac{11}{9}$