Some notations.
For any real number $x$, let's define the quantity
$$\mu(x):=\inf\left\{\mu\in\mathbb R_+\, \text{there is an infinity of rationals $p/q$ such that}\ \left\vert x-\frac pq\right\vert<\frac 1{q^{\mu}}\right\},$$
and let's call it the irrationality measure of $x$.
The problem.
Let $\lambda\in[2,+\infty[$, does there exist $x\in\mathbb R$ such that $\mu(x)=\lambda$?
The question.
I have heard that this problem has been solved, but I don't find any references for it. Do you know any (books, articles, ...)?
I think we can leverage the Sondow result from 2004 to show this (see https://arxiv.org/pdf/math/0406300.pdf for the paper, and https://mathworld.wolfram.com/IrrationalityMeasure.html for related information).
We just need to construct a simple continued fraction for which the limit supremum of the ratio of the logarithms of the denominators of two consecutive convergents is equal to $1 + \lambda$. That means that we need to select that positive integer that brings that ratio closest to $1 + \lambda$ at each step.