Solve the inequality $x^{(\log_2x)+2}\le8$

Question

The main question is:

$x^{(\log_2x)+2}\le8$. Solve the inequality.

My approach:

We can write the exponential term as :

$$\log_2(4x)$$

I actually can't go any further guys. Nothing clicks. Please help.

Answer 1

Taking logarithm wrt base $2,$

$$(\log_2x+2)\log_2x\le\log_28=3$$

Write $\log_2x=a$ $$a^2+2a-3\le 0\iff(a+3)(a-1)\le 0\iff-3\le a\le 1$$

See also: Laws of Logarithms

Answer 2

The inequality is

$$x^2(x^{\log_2 x})<8$$

Let $x=2^t$. Then the inequality becomes

$$2^{2t}(2^t)^t<8\iff 2^{2t+t^2}<2^3$$

which is equivalent to $2t+t^2<3$.