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How can we solve $e^x-e=\ln x$?

I have tried to subtract exponents, but I can't find the solution. Please help. I know there is a simple method to solve this, but I can't remember it.

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I don't think there exists a closed form solution for $x$ in terms of elementary functions. However, you can use a numerical method.

I will use the Newton-Raphson method.

The process is as follows:

$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)} \tag{1}$$

Our function to consider is given by: $$f(x)=e^x-e-\ln(x)$$ Evaluating its derivative gives: $$f'(x)=e^x-\frac{1}{x}$$ Substituting into $(1)$ gives: $$x_{n+1}=x_n-\frac{e^{x_n}-e-\ln(x_n)}{e^{x_n}-\frac{1}{x_n}}$$ Applying this iteration by setting appropriate initial values of $x_0$ (An 'educated guess' for the solution), we notice that there exists 2 solutions. As the iterations $n\to \infty$, we see that $x_n\to x$.

By setting $x_0=0.1$, we see the following:

\begin{array}{c|c}n&x_n\\\hline0&0.1\\1&0.177514\\2&0.223620\\3&0.232971\\4&0.233269\\5&0.233270\\6&0.233270\\\end{array}

After many more iterations, we obtain:

$$x \approx 0.2332697479983673$$

There is an obvious solution which is $x=1$. This solution can also be found using the same method. For example, if you set $x_0=1.5$, you will see that the iterations will converge towards $1$.

You can apply this iteration using a spreadsheet or by using more sophisticated software such as MATLAB.

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obviously is $x=1$ one solution the other one $$x \approx 0.233269747998367328501$$)we get by a numeric method, e.g. the Newton-Raphson method.

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