What are x values?log_x_2 + log_2x_2= log_4x_2 x,2x,4x are bases of logarithms

First, we'll impose the constraints of existence of
logarithms:

x>0 and x different from
1

2x>0 and x different from
1/2

4x > 0 and x different from
1/4

Now, we'll solve the equation interchanging the bases
and arguments at each of the terms of the given
equation.

1/log2 x + 1/log2 2x = 1/log2
4x

We'll apply the product property of
logarithms:

log2 2x = log2 2 + log2
x

But log2 2 = 1

log2 2x = 1 +
log2 x

log2 4x = log2 4 + log2
x

log2 4x = log2 2^2 + log2
x

log2 4x = 2log2 2 + log2
x

log2 4x = 2 + log2 x

We'll
re-write the equation:

1/log2 x + 1/(1 + log2 x) = 1/(2 +
log2 x)

We notice the presence of log2 x all over the
equation, so, we'll substitute it by t.

1/t + 1/(1+t)  =
1/(2+t)

We'll multiply each terms by
t(1+t)(2+t):

t(1+t)(2+t)/t + t(1+t)(2+t)/(1+t) =
t(1+t)(2+t)/(2+t)

We'll simplify and we'll
get:

(1+t)(2+t) + t(2+t) =
t(1+t)

We'll remove the brackets using
FOIL:

2 + 3t + t^2 + 2t + t^2 = t +
t^2

We'll subtract t + t^2:

2
+ 3t + t^2 + 2t + t^2 - t - t^2 = 0

We'll combine like
terms:

t^2 + 4t + 2 = 0

We'll
apply quadratic formula:

t1 = [-4+sqrt(16 -
8)]/2

t1 = -2+sqrt2

t2 =
-2-sqrt2

But log2 x = t

log2 x
= -2+sqrt2

x1 =
2^(-2+sqrt2)

x2 =
1/2^(2+sqrt2)

The solutions of the equation
are: x1 = 2^(-2+sqrt2) and x2 = 1/2^(2+sqrt2).