Question

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

Answer 1

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)`