1 + log₂x - 2 logₓ2 > 0

1 + log₂x - 2 (1/log₂x) > 0

Use the substitution u = log₂x ,

1 + u - 2/u > 0

u/u + u²/u - 2/u > 0

(u + u² - 2) / u > 0

(u + 2)(u - 1) / u > 0

Check for values of u to the left and right of -2 and 1 and see if they satisfy > 0 .

Note that u ≠ 0 as the denominator cannot be 0, i.e division by zero is undefined.

Result :

-2 < u < 0 or u > 1


Substitute back,

-2 < log₂x < 0 or log₂x > 1

Take log on the integers,

-2log₂2 < log₂x < 0(log₂2) or log₂x > log₂2

log₂2⁻² < log₂x < log₂2⁰ or log₂x > log₂2

2⁻² < x < 2⁰ or x > 2

¼ < x < 1 or x > 2

Thank you!

Noted the correct answer from J. Thanks