Question

Problem attached below.

Answer 1

if you look at all your answers...one sticks out of them. perhaps this will help

\[log(a+b)=log(a)+log(b)\]
\[log(-a+b)=log(b-a)=\frac{log(b)}{log(a)}\]

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

The first and last are the only two.

Answer 6

(By my above solution)

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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.

Answer 11

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

Answer 12

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