3
$\begingroup$

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?

  • 0
    thanks for all the answers (: now it's quite more clear.. but since I do not yet understand the imaginary part is more difficult..2011-11-14

1 Answers 1

8

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.