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6-log-16-x-4-3-2log-16-x-2-lt-2-

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Question Number 84674 by jagoll last updated on 15/Mar/20
((6−log_(16) (x^4 ))/(3+2log_(16) (x^2 ))) < 2
$$\frac{\mathrm{6}−\mathrm{log}_{\mathrm{16}} \:\left(\mathrm{x}^{\mathrm{4}} \right)}{\mathrm{3}+\mathrm{2log}_{\mathrm{16}} \left(\mathrm{x}^{\mathrm{2}} \right)}\:<\:\mathrm{2} \\ $$
Commented by jagoll last updated on 15/Mar/20
(i) x ≠ 0 ⇒ ((6−2log_(16) (x^2 ))/(3+2log_(16) (x^2 ))) < 2 ⇒ ((3−log_(16) (x^2 ))/(3+2log_(16) (x^2 ))) < 1 let t = log_(16) (x^2 ) ((3−t)/(3+2t))−((3+2t)/(3+2t)) < 0 ((−3t)/(3+2t)) <0 ⇒ t<−(3/2) ∨t > 0 log_(16) (x^2 ) < log_(16) (16)^(−(3/2)) ∨ log_(16) (x^2 ) > log_(16) (16)^0 x^2 < (1/(64)) ∨ x^2 > 1⇒ x<−1 ∨ −(1/8)<x<(1/8) ∨ x > 1
$$\left(\mathrm{i}\right)\:\mathrm{x}\:\neq\:\mathrm{0}\: \\ $$$$\Rightarrow\:\frac{\mathrm{6}−\mathrm{2log}_{\mathrm{16}} \:\left(\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{3}+\mathrm{2log}_{\mathrm{16}} \left(\mathrm{x}^{\mathrm{2}} \right)}\:<\:\mathrm{2} \\ $$$$\Rightarrow\:\frac{\mathrm{3}−\mathrm{log}_{\mathrm{16}} \left(\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{3}+\mathrm{2log}_{\mathrm{16}} \left(\mathrm{x}^{\mathrm{2}} \right)}\:<\:\mathrm{1}\: \\ $$$$\mathrm{let}\:\mathrm{t}\:=\:\mathrm{log}_{\mathrm{16}} \left(\mathrm{x}^{\mathrm{2}} \right) \\ $$$$\frac{\mathrm{3}−\mathrm{t}}{\mathrm{3}+\mathrm{2t}}−\frac{\mathrm{3}+\mathrm{2t}}{\mathrm{3}+\mathrm{2t}}\:<\:\mathrm{0}\: \\ $$$$\frac{−\mathrm{3t}}{\mathrm{3}+\mathrm{2t}}\:<\mathrm{0}\:\Rightarrow\:\mathrm{t}<−\frac{\mathrm{3}}{\mathrm{2}}\:\vee\mathrm{t}\:>\:\mathrm{0} \\ $$$$\mathrm{log}_{\mathrm{16}} \:\left(\mathrm{x}^{\mathrm{2}} \right)\:<\:\mathrm{log}_{\mathrm{16}} \left(\mathrm{16}\right)^{−\frac{\mathrm{3}}{\mathrm{2}}} \:\vee \\ $$$$\mathrm{log}_{\mathrm{16}} \:\left(\mathrm{x}^{\mathrm{2}} \right)\:>\:\mathrm{log}_{\mathrm{16}} \:\left(\mathrm{16}\right)^{\mathrm{0}} \\ $$$$\mathrm{x}^{\mathrm{2}} \:\:<\:\frac{\mathrm{1}}{\mathrm{64}}\:\vee\:\mathrm{x}^{\mathrm{2}} \:>\:\mathrm{1}\Rightarrow\:\mathrm{x}<−\mathrm{1}\:\vee\: \\ $$$$−\frac{\mathrm{1}}{\mathrm{8}}<\mathrm{x}<\frac{\mathrm{1}}{\mathrm{8}}\:\vee\:\mathrm{x}\:>\:\mathrm{1} \\ $$$$ \\ $$$$ \\ $$

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