TITLE: Simple math Problem. Whats wrong?
QUESTION [6 upvotes]: 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 ? :(

REPLY [10 votes]: 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 $$ e^{a-b} = \frac{e^a}{e^b} .$$

REPLY [5 votes]: 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.