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Question Number 4559 by love math last updated on 07/Feb/16

(0;(4))^(x^2 −1) >(0,(6))^(x^2 +6)

$$\left(\mathrm{0};\left(\mathrm{4}\right)\right)^{{x}^{\mathrm{2}} −\mathrm{1}} >\left(\mathrm{0},\left(\mathrm{6}\right)\right)^{{x}^{\mathrm{2}} +\mathrm{6}} \\ $$

Commented byYozzii last updated on 07/Feb/16

What is (0,(6)) and (0;(4))?

$${What}\:{is}\:\left(\mathrm{0},\left(\mathrm{6}\right)\right)\:{and}\:\left(\mathrm{0};\left(\mathrm{4}\right)\right)? \\ $$

Commented byYozzii last updated on 07/Feb/16

(0.4)^(x^2 −1) >(0.6)^(x^2 +6)   ⇒(x^2 −1)ln0.4>(x^2 +6)ln0.6  x^2 (ln0.4−ln0.6)>ln0.4+6ln0.6  x^2 ln(2/3)>ln0.4+6ln0.6  x^2 <((ln0.4+6ln0.6)/(ln(2/3)))   ⇒∣x∣<(√((ln0.4+6ln0.6)/(ln(2/3))))

$$\left(\mathrm{0}.\mathrm{4}\right)^{{x}^{\mathrm{2}} −\mathrm{1}} >\left(\mathrm{0}.\mathrm{6}\right)^{{x}^{\mathrm{2}} +\mathrm{6}} \\ $$ $$\Rightarrow\left({x}^{\mathrm{2}} −\mathrm{1}\right){ln}\mathrm{0}.\mathrm{4}>\left({x}^{\mathrm{2}} +\mathrm{6}\right){ln}\mathrm{0}.\mathrm{6} \\ $$ $${x}^{\mathrm{2}} \left({ln}\mathrm{0}.\mathrm{4}−{ln}\mathrm{0}.\mathrm{6}\right)>{ln}\mathrm{0}.\mathrm{4}+\mathrm{6}{ln}\mathrm{0}.\mathrm{6} \\ $$ $${x}^{\mathrm{2}} {ln}\left(\mathrm{2}/\mathrm{3}\right)>{ln}\mathrm{0}.\mathrm{4}+\mathrm{6}{ln}\mathrm{0}.\mathrm{6} \\ $$ $${x}^{\mathrm{2}} <\frac{{ln}\mathrm{0}.\mathrm{4}+\mathrm{6}{ln}\mathrm{0}.\mathrm{6}}{{ln}\left(\mathrm{2}/\mathrm{3}\right)}\: \\ $$ $$\Rightarrow\mid{x}\mid<\sqrt{\frac{{ln}\mathrm{0}.\mathrm{4}+\mathrm{6}{ln}\mathrm{0}.\mathrm{6}}{{ln}\left(\mathrm{2}/\mathrm{3}\right)}} \\ $$

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