(log 2 x^2)^2 + log 2 x -18 = 0

The equation `(log_2 x^2)^2 + log_2 x -18 = 0` has to be
solved

`(log_2 x^2)^2 + log_2 x -18 =
0`

Use the property of logarithm, log a^b = b*log a. This
gives:

`(2*log_2 x)^2 + log_2 x -18 =
0`

Now take the square of the first
term

`2^2*(log _2x)^2 + log_2 x -18 =
0`

Let `y = log_ 2x`

4y^2 + y
- 18 = 0

4y^2 + 9y - 8y - 18 =
0

y(4y + 9) - 2(4y + 9) = 0

(y
- 2)(4y + 9) = 0

y = 2 and y =
-9/4

Now `y = log_2 x`

`log_2
x = 2`

x = 2^2 = 4

`log_2 x =
-9/4`

x = 2^(-9/4) which is approximately
0.21022

The solution of the equation is x = 4 and x =
0.21022