## orbie 2 years ago Problem attached below.

1. orbie

2. Outkast3r09

if you look at all your answers...one sticks out of them. perhaps this will help $log(a+b)=log(a)+log(b)$ $log(-a+b)=log(b-a)=\frac{log(b)}{log(a)}$

3. LolWolf

The only solutions to the system are: $\log(a+b)=\log(a)+\log(b)\implies\\ b=\frac{a}{a-1}, a\ne0, a-1\ne0$

4. LolWolf

@Outkast3r09 $$\log(a+b)=\log(a)+\log(b)$$ Is *not* true for all cases.

5. orbie

I understand the equations but I still can't find the right answers.

6. LolWolf

The first and last are the only two.

7. LolWolf

(By my above solution)

8. orbie

mmm...that is not right either according to my math hw program. This problem is such a hassle.

9. LolWolf

Oh, I forgot to say that $$x>0$$. The last two and the first are counter-examples to these.

10. LolWolf

No, wait, never mind, it's only the first and the last.

11. LolWolf

Because that restriction doesn't hold. I don't really know... you can check them all by hand, I guess, but I'm pretty sure it's only the first and last.

12. orbie

Thank you! It was the first and the last two that worked.

13. LolWolf

Haha, yeah, I know why, now: the last one is imaginary, it *holds* but not in the real numbers.