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

Feb 25, 2016

$x = \sqrt{14}$

#### Explanation:

Simplify the left hand side through the logarithm rule:

$\log \left(a\right) + \log \left(b\right) = \log \left(a b\right)$

Thus, we obtain

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

Distributed, this gives

$\log \left({x}^{2} - 4\right) = 1$

Now, recall that $\log \left(a\right)$ really means ${\log}_{10} \left(a\right)$:

${\log}_{10} \left({x}^{2} - 4\right) = 1$

To undo the logarithm, exponentiate both sides with base $10 :$

${10}^{{\log}_{10} \left({x}^{2} - 4\right)} = {10}^{1}$

${x}^{2} - 4 = 10$

Solve:

${x}^{2} = 14$

$x = \pm \sqrt{14}$

Be very careful when solving logarithm functions--always plug your answer back into the original expression.

Note that the solution $x = - \sqrt{14}$ is invalid, since it would make $\log \left(x - 2\right)$ have a negative argument, and $\log \left(a\right)$ only exists from $a > 0$.

Thus, the only valid solution is $x = \sqrt{14}$.