# How do you solve log(x-1)+log(x+1)=2 log(x+2)?

Jan 19, 2016

There are no solutions.

#### Explanation:

Use the logarithm rules to simplify either side:

• Left hand side: $\log a + \log b = \log \left(a b\right)$
• Right hand side: $b \log a = \log \left({a}^{b}\right)$

This gives

$\log \left[\left(x - 1\right) \left(x + 1\right)\right] = \log \left[{\left(x + 2\right)}^{2}\right]$

This can be simplified using the following rule:

• If $\log a = \log b$, then $a = b$

Giving us:

$\left(x - 1\right) \left(x + 1\right) = {\left(x + 2\right)}^{2}$

Distribute both of these.

${x}^{2} - 1 = {x}^{2} + 4 x + 4$

Solve. The ${x}^{2}$ terms will cancel, so there will only be one solution.

$4 x = - 5$

$x = - \frac{5}{4}$

However, this solution is invalid. Imagine if $x$ actually were $- \frac{5}{4}$. Plug it into the original equation. The terms $\log \left(x - 1\right)$ and $\log \left(x + 1\right)$ would be $\log \left(- \frac{9}{4}\right)$ and $\log \left(- \frac{1}{4}\right)$, and the logarithm function $\log a$ is only defined when $a > 0$.