Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 100409 by Algoritm last updated on 26/Jun/20

Commented by bramlex last updated on 26/Jun/20

(((3/2)×4^x )/(4^x −9^x )) > 3+ ((4/9))^x ((3.((4/9))^x )/(2{((4/9))^x −1})) > 3+((4/9))^x let ((4/9))^x = z > 0 _(1) ((3z)/(2(z−1))) > 3+z ⇒((3z−(3+z)(z−1))/(2(z−1))) > 0 ((3z−(z^2 +2z−3))/(z−1)) > 0 ((z^2 −z−3)/(z−1)) < 0 ⇒(((z−(1/2))^2 −((13)/4))/(z−1)) < 0 (((z−(1/2)−((√(13))/2))(z−(1/2)+((√(13))/2)))/(z−1)) < 0 z < ((1−(√(13)))/2) or 1< z < ((1+(√(13)))/2) _(2) solution for z : (1)∩(2) 1 < z < ((1+(√(13)))/2) ⇒ 1< ((4/9))^x <((1+(√(13)))/2) log _(((4/9))) (((1+(√(13)))/2)) < x < 0 −1.0285 < x < 0

$$\frac{\frac{\mathrm{3}}{\mathrm{2}}×\mathrm{4}^{{x}} }{\mathrm{4}^{{x}} −\mathrm{9}^{{x}} }\:>\:\mathrm{3}+\:\left(\frac{\mathrm{4}}{\mathrm{9}}\right)^{{x}} \\ $$$$\frac{\mathrm{3}.\left(\frac{\mathrm{4}}{\mathrm{9}}\right)^{{x}} }{\mathrm{2}\left\{\left(\frac{\mathrm{4}}{\mathrm{9}}\right)^{{x}} −\mathrm{1}\right\}}\:>\:\mathrm{3}+\left(\frac{\mathrm{4}}{\mathrm{9}}\right)^{{x}} \\ $$$${let}\:\left(\frac{\mathrm{4}}{\mathrm{9}}\right)^{{x}} =\:{z}\:>\:\mathrm{0}\:\_\left(\mathrm{1}\right) \\ $$$$\frac{\mathrm{3}{z}}{\mathrm{2}\left({z}−\mathrm{1}\right)}\:>\:\mathrm{3}+{z}\:\Rightarrow\frac{\mathrm{3}{z}−\left(\mathrm{3}+{z}\right)\left({z}−\mathrm{1}\right)}{\mathrm{2}\left({z}−\mathrm{1}\right)}\:>\:\mathrm{0} \\ $$$$\frac{\mathrm{3}{z}−\left({z}^{\mathrm{2}} +\mathrm{2}{z}−\mathrm{3}\right)}{{z}−\mathrm{1}}\:>\:\mathrm{0} \\ $$$$\frac{{z}^{\mathrm{2}} −{z}−\mathrm{3}}{{z}−\mathrm{1}}\:<\:\mathrm{0}\:\Rightarrow\frac{\left({z}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{13}}{\mathrm{4}}}{{z}−\mathrm{1}}\:<\:\mathrm{0} \\ $$$$\frac{\left({z}−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{13}}}{\mathrm{2}}\right)\left({z}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{13}}}{\mathrm{2}}\right)}{{z}−\mathrm{1}}\:<\:\mathrm{0} \\ $$$${z}\:<\:\frac{\mathrm{1}−\sqrt{\mathrm{13}}}{\mathrm{2}}\:{or}\:\mathrm{1}<\:{z}\:<\:\frac{\mathrm{1}+\sqrt{\mathrm{13}}}{\mathrm{2}}\:\:\_\left(\mathrm{2}\right) \\ $$$${solution}\:{for}\:{z}\::\:\left(\mathrm{1}\right)\cap\left(\mathrm{2}\right) \\ $$$$\mathrm{1}\:<\:{z}\:<\:\frac{\mathrm{1}+\sqrt{\mathrm{13}}}{\mathrm{2}}\:\Rightarrow\:\mathrm{1}<\:\left(\frac{\mathrm{4}}{\mathrm{9}}\right)^{{x}} <\frac{\mathrm{1}+\sqrt{\mathrm{13}}}{\mathrm{2}} \\ $$$$\mathrm{log}\:_{\left(\frac{\mathrm{4}}{\mathrm{9}}\right)} \left(\frac{\mathrm{1}+\sqrt{\mathrm{13}}}{\mathrm{2}}\right)\:<\:{x}\:<\:\mathrm{0}\: \\ $$$$−\mathrm{1}.\mathrm{0285}\:<\:{x}\:<\:\mathrm{0}\: \\ $$

Commented by Algoritm last updated on 26/Jun/20

$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com