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?
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.