I am trying to find all real x satisfying the equation log(1-x) = log(1+x). I am trying to find all real $x$ satisfying the equation $\log(1-x) = \log(1+x)$. I have tried raising both sides with $e$, but I am having a hard time proving my results. Any hints? Thanks for the help!
 A: $\log$ is a one-to-one function.  Therefore $\log(1-x)=\log(1+x)$ only if $1-x=1+x$.  So take
$$
1-x = 1+x
$$
and add $x$ to both sides, and subtract $1$ from both sides.  The rest is simple.
A: I believe you are struggling with the raising both sides with $e$-part, so let me show you. You start out with:
$$\log(1-x) = \log(1+x).$$
Raising $e$ to both sides gives you
$$e^{\log(1-x)} = e^{\log(1+x)}.$$
Now, the $e$ and the $\log$ cancel each other: It is true that $e^{\log(a)} = a$ for all positive $a$. So I get $e^{\log(1-x)} = 1-x$, and $e^{\log(1+x)} = 1+x$. This means that my equation becomes
$$1-x = 1+x.$$
You can probably take it from here.
A: First notice that we need $1-x>0$ and $1+x>0$, i.e. $-1 < x < 1$.
Applying the standard "laws of logs" we have:
\begin{eqnarray*}
\ln(1-x)&=&\ln(1+x) \\ \\
\ln(1-x) - \ln(1+x) &=& 0 \\ \\
\ln\left(\frac{1-x}{1+x}\right) &=& 0 \\ \\
\frac{1-x}{1+x} &=& \mathrm{e}^0 \\ \\
\frac{1-x}{1+x} &=& 1 \\ \\
1-x &=& 1+x \\ \\
-2x &=& 0 \\ \\ 
x &=& 0
\end{eqnarray*}
A: Notice, $\log(1-x)$ is defined for all $1-x>0\iff x<1$ & 
$\log(1+x)$ is defined for all $1+x>0\iff x>-1$ 
hence, equality will hold for all real $x$ such that $$-1<x<1$$
since, logarithm is one-to-one function hence we can compare the numbers on the same base on both the sides 
$$\log(1-x)=\log(1+x)$$ 
$$\iff 1-x=1+x$$
$$2x=0$$$$ x=\color{red}{0}$$
Obviously the above value $x=0$ satisfy $-1<x<1$ hence $x=0$ is the solution
