## orbie Group Title Problem attached below. one year ago one year ago

1. orbie Group Title

2. Outkast3r09 Group Title

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 Group Title

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 Group Title

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

5. orbie Group Title

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

6. LolWolf Group Title

The first and last are the only two.

7. LolWolf Group Title

(By my above solution)

8. orbie Group Title

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

9. LolWolf Group Title

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

10. LolWolf Group Title

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

11. LolWolf Group Title

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 Group Title

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

13. LolWolf Group Title

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