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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.2011-11-14
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    that's the problem! in my calculator it has zeros... on my hp48G shows up 2.24!2011-11-14
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    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?2011-11-14
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    Looking at this link: http://www.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!2011-11-14
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    @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.2011-11-14
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    That is a nice picture! It's interesting that $f(x) = Re(\ln \frac{(x+1)}{(x-2)})$ has a solution at $x = 0.5$, though that's rather meaningless w.r.t. the actual question :)2011-11-14
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    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

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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.