Question 153463
Use the following logarithmic property:
{{{log a-log b=log (a/b)}}}
So
{{{log ((x^2-1))-log ((x-1))}}}
={{{log ((x^2-1)/(x-1))}}}
={{{log ((x+1)(x-1)/(x-1))}}}
={{{log ((x+1)cross((x-1))/cross((x-1)))}}}
={{{log ((x+1))}}}