How to find x in $x^{\log _{2}x}>16$

Question

$$x^{\log _{2} x}>16$$ What I have done is :

1. Take log fo both sides
2. Then I don't know to do what! Please help me if it is possible. Hint me about path through solving it.

Answer 1

Taking $\log_2$ on both sides, we get:

$$\log_2(x^{\log_2(x)})>\log_2(16)$$ By the power rule: $$\log(a^b)=b\log(a)$$ we get:

$$(\log_2(x))^2>\log_2(16)$$

Evaluate $\log_2(16)$ and go from there.

Answer 2

Hint: Observe \begin{align} x^{\log_2 x} = 2^{(\log_2 x)^2}. \end{align}

Answer 3

If $\log_2x=y, x=2^y$

We need $(2^y)^y>16\iff2^{y^2}>2^4\implies y^2>4$

$\implies(i)$ either $y>2\iff x>2^2$

$(ii)$ or $y<-2\iff x<2^{-2}$