The solution set of the inequation 1/(log_2x)<1/(log_2√(x+2)), is

Two sides of the given inequation are defined for $x>0$ and $x ≠ 1$.

Now, two cases arise.

CASE I When $x>1$

In this case, we have

$\log_2x>0$ and $\log_2\sqrt{x+2}>0$

$∴\frac{1}{\log_2x}<\frac{1}{\log_2\sqrt{x+2}}$

$⇒\log_2\sqrt{x+2}<\log_2x$

$⇒\sqrt{x+2}<x$

$⇒x+2<x^2$

$⇒x^2-x-2>0$

$⇒(x-2)(x+1)>0$

$⇒x-2>0$   $[∵x>1⇒x+1>0]$

$⇒x>2⇒x∈(2,∞)$

CASE II When $0<x<1$

In this case, we have

$\log_2x<0$ and $\log_2\sqrt{x+2}>0$

So, the inequation $\frac{1}{\log_2x}<\frac{1}{\log_2\sqrt{x+2}}$ holds for all $x ∈(0, 1)$.

Hence, the solution set of the given inequation is $(0,1) ∪ (2,∞)$.