I have to solve : $0 = \ln(x) - \ln(1-x).$
I did:
$0 = \ln(x) - \ln(1-x) \implies e^0 = e^{\ln(x)} - e^{\ln(1-x)} \implies 1 = x - (1-x) \implies 2 = 2x \implies x=1 $
The solution should be $x=\frac{1}{2}$
Where is my fault ? :(
I have to solve : $0 = \ln(x) - \ln(1-x).$
I did:
$0 = \ln(x) - \ln(1-x) \implies e^0 = e^{\ln(x)} - e^{\ln(1-x)} \implies 1 = x - (1-x) \implies 2 = 2x \implies x=1 $
The solution should be $x=\frac{1}{2}$
Where is my fault ? :(
After you take '$e$' of both sides you do indeed arrive at $ e^0 = e^{ \ln x - \ln(1-x) } $ because of the general property that $ a=b \implies e^a=e^b$. However, you then tried to use a false rule, that $e^{a-b} = e^{a} - e^{b} $. If you recall your index laws, you'll remember that the proper rule is $ e^{a-b} = \frac{e^a}{e^b} .$
The problem is that if you take exponentials of both sides, you should get $e^0 = e^{\ln(x) - \ln(1-x)}.$ But $e^{\ln(x) - \ln(1-x)} \neq e^{\ln(x)} - e^{\ln(1-x)}.$ So your first step is incorrect.
Simplest thing is to first go from $0 = \ln(x) - \ln(1-x)$ to $\ln(1-x) = \ln(x)$ and then take the exponentials.
Alternatively, from $ e^0 = e^{\ln (x) - \ln(1-x)}$ you can use the fact that $e^{a-b} = \frac{e^a}{e^b}$ to get $ 1 = \frac{e^{\ln(x)}}{e^{\ln(1-x)}},$ and proceed from there.
P.S. Please don't use [homework] as the only tag in your post.