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

##### 1 Answer

It is

$\log \left(x + 1\right) + \log \left(x - 1\right) = 1 \implies \log \left({x}^{2} - 1\right) = \log 10 \implies {x}^{2} - 1 = 10 \implies {x}^{2} = 11 \implies x = \sqrt{11}$

If log represents the natural logarithm we have that

$\log \left({x}^{2} - 1\right) = 1 \implies \log \left({x}^{2} - 1\right) = \log e \implies x = \sqrt{1 + e}$