Solving $2\log_2 (\log_2 x) + \log_{1/2} (\log_2 x) = 1$ just looking for some help with the following:
$$2\log_2 (\log_2 x) + \log_{1/2} (\log_2 x) = 1$$
I know how to solve a log equation with a single nested log that's $= 1$ or $0$ or some number, but I'm unsure of what to do when summing/ subtracting nested logs like these.
If possible, a hint would be much more appreciated than the flat out answer.
Thanks in advance to anyone who stops to help.
 A: 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 $1/2 = 2^{-1}$)
A: $1)$ 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.
A: 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:
$$x=4\tag{13}$$

