algebra precalculus - What are the solutions for the following equation?

I have the following equation:

$\log_{2x}4x+\log_{4x}16x=4$

What are the solutions of this equation?

This is what I did:

Firstly, I applied the following conditions:

$2x>0 \Rightarrow x>0$

$2x \ne 1 \Rightarrow x \ne \dfrac{1}{2}$

$4x>0 \Rightarrow x>0$

$4x \ne 1 \Rightarrow x \ne \dfrac{1}{4}$

$16x > 0 \Rightarrow x > 0$

Considering all of these at once, I have: $x \in (0, + \infty) \setminus \bigg\{\dfrac{1}{2}, \dfrac{1}{4} \bigg \}$

Next, I wrote the equation like so:

$\log_{2x}(2x*2) + \log_{4x}(4x*4)=4$

$\log_{2x}2x + \log_{2x}2 + \log_{4x}4x+\log_{4x}4 = 4$

$1 + \log_{2x}2 + 1 + \log_{4x}4 = 4$

$\log_{2x}2 + \log_{4x}4 =2$

$\log_{2x}2 + 2\log_{4x}2=2$

$\log_{2x}2 + \dfrac{2}{\log_{2}4x}=2$

$\log_{2x}2 + \dfrac{2}{\log_2 2 + \log_2 2x}=2$

$\log_{2x}2+\dfrac{2}{1+\dfrac{1}{\log_{2x}2}}=2$

Then I used the substitution $\log_{2x}2=t$

$t+ \dfrac2{1+ \dfrac{1}{t}}=2$

$...$

$t^2+t-2=0$

$(t+2)(t-1)=0$

In the case that $t=-2$,

$\log_{2x}2=-2$

$(2x)^{-2}=2$

$\dfrac{1}{4x^2}=2$

$x^2= \dfrac{1}{8}$

$x = \pm \dfrac{1}{2 \sqrt{2}}$, but because of our conditions: $x = \dfrac{1}{2 \sqrt{2}}$

And in the case that $t=1$,

$\log_{2x}2=1$

$2x=2$

$x=1$

So, in the end I got: $x \in \bigg \{ \dfrac{1}{2 \sqrt{2}}, 1 \bigg \}$. The only problem with this is that it is completly wrong...

My textbook says that the correct answer should be: $\bigg [ \dfrac{1}{2 \sqrt{2}}, 2 \bigg ]$. So you see I'm not even close, I need an interval of values, not just $2$ solutions like I got. So, where did I make mistakes and what should I do to get the right answer?

Answer

As mentioned in the comments, your answer is perfectly correct and the textbook is the one which is wrong. You can check this by substituting the values in. Another way to see that it certainly cannot be an interval of solutions is by noticing that the equation $$\log_{2x}2+\log_{4x}4=2,\tag{1}$$
which is equivalent to the given one, has a LHS which changes with $x$, and on no interval $[a,b]$ is it constant. So it is absurd that there can be an interval on which the LHS is constant at $2$.

I will say though that in questions involving the logarithm like this one, a smart general approach is to convert everything to the same base. So starting from $(1)$, write
$$\frac{1}{1+\log_2(x)}+\frac{2}{2+\log_2(x)}=2,$$
and then substitute $y=\log_2(x)$ to solve for $y$ and hence for $x$. This is slightly cleaner than your solution, but of course yours is correct too.