Solving $2\log_2 (\log_2 x) + \log_{1/2} (\log_2 x) = 1$

I would first perform a change of basis in the second term.

Recall: $\log_{(1/2)}a = \frac{\log_2(a)}{\log_2(1/2)}$

(Of course, then see that $1/2 = 2^{-1}$)

$1)$ Use change of base, to write $$\log_{1/2}(\log_2 x) = - \log_2(\log_2x) $$

$2)$ Let $$\log_2(\log_2 x) = t$$

$3)$ Then $$t= ?$$

$4)$ This means $$\log_2 x =?$$

$5)$ Finally, this means $$x=?$$

$6)$ Check this root in the original equation.

Well, let's solve a more general problem:

$$\text{n}\log_\alpha\left(\log_\alpha\left(x\right)\right)+\log_\beta\left(\log_\alpha\left(x\right)\right)=\epsilon\tag1$$

First, we know that:

$$\log_\text{a}\left(\text{b}\right):=\frac{\ln\left(\text{b}\right)}{\ln\left(\text{a}\right)}\tag2$$

Let $\text{p}:=\log_\alpha\left(x\right)$, so we get for $(1)$:

$$\text{n}\log_\alpha\left(\text{p}\right)+\log_\beta\left(\text{p}\right)=\epsilon\tag3$$

Using $(2)$ we get:

$$\text{n}\cdot\frac{\ln\left(\text{p}\right)}{\ln\left(\alpha\right)}+\frac{\ln\left(\text{p}\right)}{\ln\left(\beta\right)}=\epsilon\tag4$$

Rewrite the left-hand side, of $(4)$ by combining fractions:

$$\frac{\ln\left(\text{p}\right)\left(\ln\left(\alpha\right)+\text{n}\ln\left(\beta\right)\right)}{\ln\left(\alpha\right)\ln\left(\beta\right)}=\epsilon\tag5$$

Solving for $\ln\left(\text{p}\right)$, gives:

$$\ln\left(\text{p}\right)=\epsilon\cdot\frac{\ln\left(\alpha\right)\ln\left(\beta\right)}{\ln\left(\alpha\right)+\text{n}\ln\left(\beta\right)}\tag6$$

Cancel logarithms by taking $\exp$ of both sides:

$$\text{p}=\exp\left(\epsilon\cdot\frac{\ln\left(\alpha\right)\ln\left(\beta\right)}{\ln\left(\alpha\right)+\text{n}\ln\left(\beta\right)}\right)\tag7$$

Setting the substitution back and using $(2)$:

$$\log_\alpha\left(x\right)=\frac{\ln\left(x\right)}{\ln\left(\alpha\right)}=\exp\left(\epsilon\cdot\frac{\ln\left(\alpha\right)\ln\left(\beta\right)}{\ln\left(\alpha\right)+\text{n}\ln\left(\beta\right)}\right)\tag8$$

Solving for $\ln\left(x\right)$, gives:

$$\ln\left(x\right)=\ln\left(\alpha\right)\exp\left(\epsilon\cdot\frac{\ln\left(\alpha\right)\ln\left(\beta\right)}{\ln\left(\alpha\right)+\text{n}\ln\left(\beta\right)}\right)\tag9$$

Cancel logarithms by taking $\exp$ of both sides:

$$x=\exp\left(\ln\left(\alpha\right)\exp\left(\epsilon\cdot\frac{\ln\left(\alpha\right)\ln\left(\beta\right)}{\ln\left(\alpha\right)+\text{n}\ln\left(\beta\right)}\right)\right)\tag{10}$$

Using:

$$\ln\left(\text{a}^\text{b}\right)=\text{b}\ln\left(\text{a}\right)\tag{11}$$

We can rewrite $(10)$ as follows:

$$x=\alpha^{\exp\left(\frac{\epsilon\ln\left(\alpha\right)\ln\left(\beta\right)}{\ln\left(\alpha\right)+\text{n}\ln\left(\beta\right)}\right)}\tag{12}$$

Using your values, we get:

$$x=4\tag{13}$$