# 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?

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Your steps are correct till now; the next crucial step is to conclude that there is no $x$ such that $\ln (\frac{x+1}{x-2}) = 0$. In other words, the equation has no solutions. –  Srivatsan Nov 14 '11 at 1:11
that's the problem! in my calculator it has zeros... on my hp48G shows up 2.24! –  Totty Nov 14 '11 at 1:14
Plugging in $x = 2.24$ in the formula for $f(x)$ gives the value $\ln (3.24 / .24) \approx 2.603$, which is quite far from $0$. Are you sure you are not making some mistake in finding the zeroes in your calculator? –  Srivatsan Nov 14 '11 at 1:17
Looking at this link: wolframalpha.com/input/?i=ln%28%28x%2B1%29%2F%28x-2%29%29 you see that at the zero of this function, there is also an imaginary part. This equation has no solution in the set of real numbers. This shows graphically why there is no solution, because when the real part is zero, then the imaginary part is nonzero! –  tomcuchta Nov 14 '11 at 1:18
@Totty What do you mean by "in my calculator"? Do you mean you graphed the function? I graphed it as well and see no roots. –  Austin Mohr Nov 14 '11 at 1:19
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.