1-x-2x-x-4x-6-x-0-x-

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

\frac{\overset{}{1}}{x^{2 x}} + x^{- 4 x} = 6, (x \neq 0)

x = \underset{}{?}

Commented by MJS last updated on 03/May/20

x^{- 2 x} + x^{- 4 x} = 6

t + t^{2} = 6

t = - 3 \lor t = 2

x^{- 2 x} = - 3 no solution

x^{- 2 x} = 2

u = \frac{1}{x}

\sqrt[u]{u^{2}} = 2 \Rightarrow u = 2 \lor u = 4 (obviously)

\Rightarrow

x = \frac{1}{4} \lor x = \frac{1}{2}

Commented by jagoll last updated on 03/May/20

multiply x^{4 x}

x^{2 x} + 1 = 6 x^{4 x} [s e t x^{2 x} = p]

6 p^{2} - p - 1 = 0

(3 p + 1) (2 p - 1) = 0

p = \frac{1}{2} \Rightarrow x^{2 x} = 2^{- 1} \Rightarrow 2 = x^{- 2 x}

{(\frac{1}{x^{x}})}^{2} = {(\sqrt{2})}^{2} \Rightarrow x^{x} = \frac{1}{\sqrt{2}}

{(\frac{1}{2})}^{\frac{1}{2}} = x^{x} o r {(\frac{1}{4})}^{\frac{1}{4}} = x^{x}

{\begin{cases} x = \frac{1}{2} \\ x = \frac{1}{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