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I need help solving this equation: $$\frac{e^x-1-(x-2)e^x}{(e^x-1)^2}=1.$$ I have tried many times but I still can't solve it...

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    Please check that I have not butchered your equation while editing. I freely replaced a big $X$ by a small $x$, I thought it was a typo. – 2017-01-30
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    Yes, it's fine :) – 2017-01-30
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    The problem is, I don't learn that method in my school.. they don't that method in my country :/ – 2017-01-30
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    Perhaps some context can help give you a better answer. In what context do you need to solve this equation? – 2017-01-30
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    Your question was put on hold, the message above (and possibly comments) should give an explanation why. (In particular, [this link](http://meta.math.stackexchange.com/a/9960) might be useful.) You might try to edit your question to address these issues. Note that the next edit puts your post in the review queue, where users can vote whether to reopen it or leave it closed. (Therefore it would be good to avoid minor edits and improve your question as much as possible with the next edit.) – 2017-01-30
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    @SarahJones Please do not vandalize your own question. – 2017-01-30

2 Answers 2

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I do not think that we could find the analytical solution for the zeros of function $$f(x)=\frac{e^x-1-(x-2)e^x}{(e^x-1)^2}-1$$ and numerical methods should be used.

Since I suppose that you already made a plot of this function, you probably noticed that the roots are close to $x=1$ and $x=-1$.

So, using these as initial guesses, apply Newton method which, starting from a "reasonable" guess, will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$

Using $x_0=1$, the iterates would be $$\left( \begin{array}{cc} n & x_n \\ 0 & 1\\ 1 & 1.131115609 \\ 2 & 1.164773954 \\ 3 & 1.166455518 \\ 4 & 1.166459408 \end{array} \right)$$

Using $x_0=-1$, the iterates would be $$\left( \begin{array}{cc} n & x_n \\ 0 & -1 \\ 1 & -1.043535838 \\ 2 & -1.046488210 \\ 3 & -1.046500386 \\ 4 & -1.046500387 \end{array} \right)$$

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We can write that as: $$\frac{e^x-1-(x-2)e^x}{(e^x-1)^2}=\Big(\frac{x-2}{e^x-1}\Big)'=1$$ this means $$\frac{x-2}{e^x-1}=x+C$$ we need a point of it to determine constant $C$ and this value specifies whether we can continue for solutions or not.