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

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Answer 1 · 2015-11-29T00:27:13

It is

$log(x+1)+log(x-1)=1=>log(x^2-1)=log10=>x^2-1=10=>x^2=11=>x=sqrt11$

If log represents the natural logarithm we have that

$log(x^2-1)=1=>log(x^2-1)=loge=>x=sqrt(1+e)$

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