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My friend is kind of stuck with this problem and I can't find a numerical solution. I plotted the graph to see that there are 2 different solutions but I don't know to mathematically find them. Can you suggest me a method at least that I can apply for this equation?

$$\ln(10x) + 1 - x = 0$$

user376343
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Alexandru Ionita
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    Did you try the newton method , the bisection method or regula falsi ? What was the problem ? – Peter Nov 01 '21 at 12:54
  • Perhaps the most common method is Newton's method: https://en.wikipedia.org/wiki/Newton%27s_method – Matti P. Nov 01 '21 at 12:54
  • How far into mathematics are you? Do you know how to take derivatives? Do you know what a differential equation is? Do you know what conditions a functions needs to satisfy to be invertible? These questions need clarification before anyone should attempt to answer your question. – Angel Nov 01 '21 at 13:02
  • In case you were using the phrase "numerical solution" to merely mean "numerical value of the solution", do note that in Mathematics, "numerical solution" has a particular meaning in contrast to "analytical solution": the former uses numerical approximation, whereas the latter uses purely symbolic manipulation. – ryang Nov 01 '21 at 13:11
  • Iterating the functions $f(x)=\ln(10x)+1$ and $g(x)=1/(\ln(10/x)+1)$ starting from $x=1$ will converge to the two solutions. – Barry Cipra Nov 01 '21 at 18:57

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$$\ln(10x) +1 - x =0$$ $$\ln(10xe^{-x}) = -1$$ $$10xe^{-x} = \frac{1}{e}$$ $$-xe^{-x} = -\frac{1}{10e}$$ Since $-\frac{1}{10e}\in[-\frac{1}{e},0)$, $-\frac{1}{10e}$ is in the domain of both branches of the $W$ function: $$-x = W_{0}(-\frac{1}{10e}) \text{ or }W_{-1}(-\frac{1}{10e})$$ $$\boxed{x = -W_{0}(-\frac{1}{10e})\text{ or }-W_{-1}(-\frac{1}{10e})}$$

Joshua Wang
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