What is the positive difference between the two real values of ‘x’ for which (log2 x)4 + 12(log2(1/x))2 – 26 = 0?

To solve this problem, we first need to simplify the equation.

Given equation is (log2 x)⁴ + 12(log2(1/x))² – 26 = 0

We know that log2(1/x) = -log2x.

So, the equation becomes (log2 x)⁴ - 12(log2x)² – 26 = 0

This equation is in the form of a quadratic equation (a²)² - 12a² - 26 = 0, where a = log2x.

Let's solve this quadratic equation:

a² = [12 ± sqrt((12)² - 41(-26))]/2*1 a² = [12 ± sqrt(144 + 104)]/2 a² = [12 ± sqrt(248)]/2 a² = [12 ± 2sqrt(62)]/2 a² = 6 ± sqrt(62)

So, the two possible values for a (log2x) are sqrt(6 + sqrt(62)) and sqrt(6 - sqrt(62)).

Therefore, the two possible values for x are 2^(sqrt(6 + sqrt(62))) and 2^(sqrt(6 - sqrt(62))).

The positive difference between these two values will be |2^(sqrt(6 + sqrt(62))) - 2^(sqrt(6 - sqrt(62)))|.