Question

How to solve `log(x-1)+log(x+1)=2log(x+2)`?

Answer 1

For `x>1` we have that

`log(x-1)+log(x+1)=log(x+2)^2`

`Log(x^2-1)=log(x+2)^2`

`x^2-1=(x+2)^2`

`x^2-1=x^2+4x+4`

`x=-5/4`

However this solution is not acceptable because we are looking solutions for `x>1`.That is because `loga` is defined for `a>0`

Hence there are NO solutions for this equation.

The following logarithm rules were used

`loga+logb=log(ab)`

`bloga=log(a^b)`

If `loga=logb`, then `a=b`