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Question Number 91870 by hmamarques1994@gmail.com last updated on 03/May/20

$$\frac{\stackrel{}{1}}{{\mathrm{x}}^{2\mathrm{x}}}+{\mathrm{x}}^{-4\mathrm{x}}=6,(\mathrm{x}\ne 0)$$$$$$$$\mathrm{x}=\underset{}{?}$$

Commented by MJS last updated on 03/May/20

$$x^{-2x}+x^{-4x}=6$$$$t+t^2=6$$$$t=-3\vee t=2$$$$x^{-2x}=-3\mathrm{no}\mathrm{solution}$$$$x^{-2x}=2$$$$\mathrm{u}=\frac{1}{x}$$$$\sqrt[u]{u^2}=2\Rightarrow u=2\vee u=4\left(\mathrm{obviously}\right)$$$$\Rightarrow$$$$x=\frac{1}{4}\vee x=\frac{1}{2}$$

Commented by jagoll last updated on 03/May/20

$$\mathrm{multiply}{x}^{4x}$$$$x^{2x}+1=6x^{4x}[set{x}^{2x}=p]$$$$6p^2-p-1=0$$$$(3p+1)(2p-1)=0$$$$p=\frac{1}{2}\Rightarrow x^{2x}=2^{-1}\Rightarrow 2=x^{-2x}$$$${\left(\frac{1}{x^x}\right)}^2={\left(\sqrt{2}\right)}^2\Rightarrow x^x=\frac{1}{\sqrt{2}}$$$${\left(\frac{1}{2}\right)}^{\frac{1}{2}}=x^{x}or{\left(\frac{1}{4}\right)}^{\frac{1}{4}}=x^x$$$$\{\begin{array}{l}x=\frac{1}{2}\\ x=\frac{1}{4}\end{array}$$$$$$

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

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