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Question Number 140589 by bramlexs22 last updated on 09/May/21

$$x^{-\log {}_{2}x+4}<\frac{x}{16}$$

Answered by MJS_new last updated on 10/May/21

$$x^{4-{\log }_{2}x}\in \mathbb{R}\Rightarrow x>0$$ $$\mathrm{let}x=2^t$$ $${\left(2^t\right)}^{4-t}<\frac{2^t}{2^4}$$ $$2^{4t-t^2}<2^{t-4}$$ $$4t-t^2<t-4$$ $$t^2-3t-4>0$$ $$\Rightarrow t<-1\vee t>4$$ $$\Rightarrow x<\frac{1}{2}\vee x>16$$ $$\Rightarrow$$ $$0<x<\frac{1}{2}\vee x>16$$

Commented bybramlexs22 last updated on 10/May/21

$$\mathrm{nice}\mathrm{sir}$$

Answered by bemath last updated on 10/May/21

$$\log {}_2\left(\mathrm{x}\right)(4-\log {}_2\left(\mathrm{x}\right))<\log {}_2\left(\mathrm{x}\right)-4$$ $$\mathrm{let}\log {}_2\left(\mathrm{x}\right)=\mathrm{b}$$ $$\Rightarrow 4\mathrm{b}-{\mathrm{b}}^2<\mathrm{b}-4$$ $$\Rightarrow {\mathrm{b}}^2-3\mathrm{b}-4>0$$ $$\Rightarrow (\mathrm{b}-4)(\mathrm{b}+1)>0$$ $$\Rightarrow \mathrm{b}<-1\vee \mathrm{b}>4$$ $$\Rightarrow 0<\mathrm{x}<\frac{1}{2}\vee \mathrm{x}>16$$

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