Wolfram Alpha gives a numerical solution to this equality, but is it not possible to find a symbolic answer? Why not? What types of functions with log in it can and can't you solve and why?
Thanks!
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2$\log(x)$ is a transcendental function, so these sorts of equations cannot be solved usually without numerical methods. – A-Level Student Jan 22 '21 at 10:22
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It is surprising (I repeated the calculation with Wolfram Alpha), that you did not receive the sybolic answer since $$\log(x)=a x+b \implies x=-\frac 1a W\left(-a e^b\right)$$ where $W(t)$ is Lambert function.
Applied to your case, this gives, beside the trivial $x=1$ $$x=-2 W_{-1}\left(-\frac{1}{2 \sqrt{e}}\right)\sim 3.51286$$
Claude Leibovici
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How do you find the $W_{-1}$ branch in this case? I ended up with $x=-2W_0(-\frac{1}{2}e^{-\frac{1}{2}})$ which is indeed the trivial $x=1$ – GambitSquared Jan 22 '21 at 18:18
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@GambitSquared.Take care with $W(.)$. It is multivalued. In the real domain, you have $W_0$ and $W_{-1}$ – Claude Leibovici Jan 22 '21 at 18:23
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Yes, but I don't see where the second value comes from... there is no quadratic involved, is there? – GambitSquared Jan 22 '21 at 18:25
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Here I have outlined my question: https://math.stackexchange.com/questions/3995827/find-the-two-branches-of-log-x-frac12x-1 – GambitSquared Jan 22 '21 at 18:39
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If $-e^{-1} < y < 0$ then $x e^x = y$ has two real solutions. (Plot the graph to see why.) One in $(-1,0)$ is given by $W_0(y)$ and one less than $-1$ is given by $W_{-1}(y)$. – WimC Jan 22 '21 at 18:56