Question Number 91870 by hmamarques1994@gmail.com last updated on 03/May/20
$$\:\frac{\overset{\:} {\mathrm{1}}}{\mathrm{x}^{\mathrm{2x}} }\:+\:\mathrm{x}^{−\mathrm{4x}} \:=\:\mathrm{6},\:\:\left(\mathrm{x}\:\neq\:\mathrm{0}\right) \\ $$$$\: \\ $$$$\:\mathrm{x}\:=\:\underset{\:} {?} \\ $$
Commented by MJS last updated on 03/May/20
$${x}^{−\mathrm{2}{x}} +{x}^{−\mathrm{4}{x}} =\mathrm{6} \\ $$$${t}+{t}^{\mathrm{2}} =\mathrm{6} \\ $$$${t}=−\mathrm{3}\vee{t}=\mathrm{2} \\ $$$${x}^{−\mathrm{2}{x}} =−\mathrm{3}\:\mathrm{no}\:\mathrm{solution} \\ $$$${x}^{−\mathrm{2}{x}} =\mathrm{2} \\ $$$$\mathrm{u}=\frac{\mathrm{1}}{{x}} \\ $$$$\sqrt[{{u}}]{{u}^{\mathrm{2}} }=\mathrm{2}\:\Rightarrow\:{u}=\mathrm{2}\vee{u}=\mathrm{4}\:\left(\mathrm{obviously}\right) \\ $$$$\Rightarrow \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{4}}\vee{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by jagoll last updated on 03/May/20
![multiply x^(4x) x^(2x) +1=6x^(4x) [ set x^(2x) = p] 6p^2 −p−1=0 (3p+1)(2p−1)=0 p= (1/2) ⇒ x^(2x) = 2^(−1) ⇒2=x^(−2x) ((1/x^x ))^2 = ((√2))^2 ⇒x^x = (1/( (√2))) ((1/2))^(1/2) = x^x or ((1/4))^(1/4) = x^x { ((x=(1/2))),((x=(1/4))) :}](https://www.tinkutara.com/question/Q91882.png)
$$\mathrm{multiply}\:{x}^{\mathrm{4}{x}} \\ $$$${x}^{\mathrm{2}{x}} +\mathrm{1}=\mathrm{6}{x}^{\mathrm{4}{x}} \:\left[\:{set}\:{x}^{\mathrm{2}{x}} =\:{p}\right] \\ $$$$\mathrm{6}{p}^{\mathrm{2}} −{p}−\mathrm{1}=\mathrm{0} \\ $$$$\left(\mathrm{3}{p}+\mathrm{1}\right)\left(\mathrm{2}{p}−\mathrm{1}\right)=\mathrm{0} \\ $$$${p}=\:\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:{x}^{\mathrm{2}{x}} =\:\mathrm{2}^{−\mathrm{1}} \Rightarrow\mathrm{2}={x}^{−\mathrm{2}{x}} \\ $$$$\left(\frac{\mathrm{1}}{{x}^{{x}} }\right)^{\mathrm{2}} =\:\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \:\Rightarrow{x}^{{x}} =\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} =\:{x}^{{x}} \:{or}\:\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} =\:{x}^{{x}} \\ $$$$\begin{cases}{{x}=\frac{\mathrm{1}}{\mathrm{2}}}\\{{x}=\frac{\mathrm{1}}{\mathrm{4}}}\end{cases} \\ $$$$ \\ $$
Commented by john santu last updated on 03/May/20
hello mister mjs
Commented by MJS last updated on 03/May/20
hi! ��
Commented by john santu last updated on 03/May/20
Look at the matter number 91786