# How do you solve ln (x+5) = ln (x-1) - ln (x+1)?

First use the fact that the difference of two logs is the log of the quotient to write $\ln \left(x + 5\right) = \ln \left(\setminus \frac{x - 1}{x + 1}\right)$. Now exponentiate both sides to get $x + 5 = \setminus \frac{x - 1}{x + 1}$. Multiplication of both sides of this equation by $x + 1$ and expansion leads to ${x}^{2} + 6 x + 5 = x - 1$, and rearrangement gives ${x}^{2} + 5 x + 6 = 0$. Factoring this results in $\left(x + 3\right) \left(x + 2\right) = 0$ so that two possible solutions are $x = - 3$ and $x = - 2$.
Solving logarithmic equations often results in extraneous solutions, so it's always good to check the answers in the original equation. In fact, in this case, neither of these answers works in the original equation because the domain of the natural logarithm function is the set of all positive real numbers. (These values of $x$ do solve the equation $\ln \left(x + 5\right) = \ln \left(\setminus \frac{x - 1}{x + 1}\right)$, however).