What is x if (log2 x)^2+log2 (4x)=4?

First, we'll use the product rule of logarithms and we'll
re-write the term log2 (4x).

log2 (4x) = log2 4 + log2
x

log2 (4x) = log2 2^2 + log2
x

We'll apply the power rule of
logarithms:

log2 (4x) = 2log2 2 + log2
x

But log2 2 = 1

log2 (4x) = 2
+ log2 x

We'll substitute the term log2 (4x) in the given
equation:

(log2 x)^2 +  2 + log2 x =
4

We'll substitute  log2 x =
t

We'll re-write the equation in
t:

t^2 + 2 + t - 4 = 0

We'll
combine like terms:

t^2 + t - 2 =
0

We'll apply the quadratic
formula:

t1 = [-1+sqrt(1 +
8)]/2

t1 = (-1+3)/2

t1 =
1

t2 = (-1-3)/2

t2 =
-2

We'll put:

log2 x =
t1

log2 x = 1

x =
2^1

x = 2

log2 x =
t2

log2 x = -2

x =
2^-2

x = 1/2^2

x =
1/4

Since both solutions are positive, we'll
accept them: {1/4 ; 2}.