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# Solve $\frac{x+\dots+x^K}{K} = \frac{1}{2}$ for large values of $K$

I am interested in the unique solution $$x$$ for the equation :

$$p_K(x)=\frac{x+\dots+x^K}{K}=\frac{1}{2},$$
for large values of $$K$$. When $$K$$ is small ($$K=1$$ and $$K=2$$) we can solve this equation explicitly and find :
$$x=\frac{1}{2}, \frac{\sqrt{5}-1}{2}.$$
For $$K=3$$ we still get an explicit solution which is more complicated and from $$K=4$$ on I do not find an explicit solution. As $$K$$ tends to infinity, we find that the unique solution $$x_K$$ of $$p_K(x_K)=0$$ tends to one, i.e. $$\lim_{K \rightarrow \infty} x_K = 1$$. I would like to find an asymptotic approximation of $$p_K(x)$$ denoted by $$\tilde p_K(x)$$ for which the solutions $$\tilde x_K$$ satisfy:
$$\lim_{K\rightarrow \infty} K \cdot \log(x_K) = \lim_{K\rightarrow \infty} K \cdot \log(\tilde x_K).$$
My idea was to use the approximation :
$$\frac{x+\dots+x^K}{K} \approx x^{\frac{\sum_{j=1}^K j }{K}} = x^{\frac{K+1}{2}}.$$
Using this approximation, we find $$\tilde x_K = \left( \frac{1}{2} \right) ^{\frac{2}{K+1}}$$ and we find for the limit :
$$\lim_{K\rightarrow \infty} K \cdot \log(\tilde x_K) = -2 \log(2) \approx -1.38$$
but by numerical approximation, find :
$$\lim_{K\rightarrow \infty} K \cdot \log(x_K) \approx – 1.592$$