## anonymous 3 years ago Problem attached below.

1. anonymous

2. anonymous

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. anonymous

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. anonymous

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

5. anonymous

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

6. anonymous

The first and last are the only two.

7. anonymous

(By my above solution)

8. anonymous

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

9. anonymous

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

10. anonymous

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

11. anonymous

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. anonymous

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

13. anonymous

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