Question

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

Answer 1

There are no solutions.

Explanation:

Use the logarithm rules to simplify either side:

• Left hand side: `loga+logb=log(ab)`
• Right hand side: `bloga=log(a^b)`

This gives

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

This can be simplified using the following rule:

• If `loga=logb`, then `a=b`

Giving us:

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

Distribute both of these.

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

Solve. The `x^2` terms will cancel, so there will only be one solution.

`4x=-5`

`x=-5/4`

However, this solution is invalid. Imagine if `x` actually were `-5/4`. Plug it into the original equation. The terms `log(x-1)` and `log(x+1)` would be `log(-9/4)` and `log(-1/4)`, and the logarithm function `loga` is only defined when `a>0`.