How to find the roots of $f(x)= \ln( \frac{x+1 }{x-2})$? I can't solve this equation:
$$\ln\left(\frac{x+1}{x-2}\right) = 0.$$
I do:
$$\begin{align*}
 \ln \left( \frac{x+1}{x-2} \right)&=0\\
 \frac{x+1}{x-2} &= 1 \\
x+1&=x-2 \\
 x+1-x+2&=0 \\
x-x+3&=0 \\
 3&=0 
\end{align*}$$
Then $x$ is? 

What you've shown is that $\frac{x+1}{x-2}$ is never equal to 1.  Since 1 is the only value where natural log equals zero, the equation $\log \frac{x+1}{x-2} = 0 $ has no solutions.
