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

Question

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

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

Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps. — robjohn, Jan 27 '19 at 18:15

score 1 · Answer 1 · answered Jan 27 '19 at 17:59

1

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

answered Jan 27 '19 at 17:59

Jacky Chong

score 1 · Accepted Answer · answered Jan 27 '19 at 18:05

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.

answered Jan 27 '19 at 18:05

Rhys Hughes

Would you explain that how you used that power rule in second step? – Mr invisible Jan 27 '19 at 18:15
I brought the power on the LHS down, so: $$\log_2(x^{\log_2(x)})=\log_2(x)\log_2(x)=(\log_2(x))^2$$ – Rhys Hughes Jan 27 '19 at 18:38

score 0 · Answer 3 · answered Jan 27 '19 at 18:23

0

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 $y<-2\iff x<2^{-2}$

answered Jan 27 '19 at 18:23

lab bhattacharjee

3 Answers3