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How to solve this equation : $$\left ( \exp \left ( x \right )-1 \right )\left ( \ln \left ( \exp \left ( x \right )-1 \right ) \right )-x\exp \left ( x \right )\ln \left ( x \right )=0$$

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  • $\begingroup$ I would begin by using exp(0)=1, ln is inverse of exp, exp(a+b)=exp(a)exp(b) $\endgroup$ – Emil Feb 26 '17 at 12:02
  • $\begingroup$ ln(xy)=ln(x)+ln(y) seems handy too. $\endgroup$ – Emil Feb 26 '17 at 12:17
  • $\begingroup$ this equation can only be solved by a numerical method,e.g. Newton-Raphson method $\endgroup$ – Dr. Sonnhard Graubner Feb 26 '17 at 13:45
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As said in comments, this is a transcendental equation which will not show analytical expression for the roots (except the trivial $x=0$) and only numerical methods, such as Newton, will be required.

So consider $$f(x)=\left(e^x-1\right) \log \left(e^x-1\right)-e^x x \log (x)$$ $$f'(x)=e^x \left(\log \left(e^x-1\right)-(x+1) \log (x)\right)$$ and use the iterative scheme $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$

Looking at the plot of the function reveals that the root is close to $2.5$. But, let me be very lazy using $x_0=3$. The successive iterates will be $$\left( \begin{array}{cc} n & x_n \\ 0 & 3 \\ 1 & 2.658444549 \\ 2 & 2.462103808 \\ 3 & 2.398181899 \\ 4 & 2.392043125 \\ 5 & 2.391990039 \\ 6 & 2.391990035 \end{array} \right)$$ which is the solution for ten significant figures.

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