x-2-2-x-Find-more-than-three-solutions-

Question Number 211675 by MrGaster last updated on 16/Sep/24

x^{2} = 2^{x}

Find more than three solutions

Commented by mr W last updated on 16/Sep/24

b u t t h e r e a r e o n l y t h r e e s o l u t i o n s!

Answered by mr W last updated on 16/Sep/24

x^{2} = 2^{x}

\Rightarrow x = \pm 2^{\frac{x}{2}}

\Rightarrow (- \frac{x}{2}) \times 2^{- \frac{x}{2}} = \pm \frac{1}{2}

\Rightarrow (- \frac{x}{2}) (\ln 2) \times e^{(- \frac{x}{2}) (\ln 2)} = \pm \frac{\ln 2}{2}

\Rightarrow (- \frac{x}{2}) (\ln 2) = W (\pm \frac{\ln 2}{2})

\Rightarrow x = - \frac{2}{\ln 2} \times W (\pm \frac{\ln 2}{2}) = {\begin{cases} 2 \\ {\begin{cases} 4 \\ - 0.766664696 \end{cases} \end{cases}

Answered by Frix last updated on 16/Sep/24

x^{2} = 2^{x}

2 \ln x = x \ln 2

- \frac{\ln x}{x} = - \frac{\ln 2}{2}

x = e^{- t}

e^{t} t + \frac{\ln 2}{2} = 0

t = a + b i

(e^{a} (a \cos b - b \sin b) + \frac{\ln 2}{2}) + e^{a} (a \sin b + b \cos b) i = 0

{\begin{cases} e^{a} (a \cos b - b \sin b) + \frac{\ln 2}{2} = 0 \\ e^{a} (a \sin b + b \cos b) = 0 \Rightarrow a = - b \cot b \end{cases}

\Rightarrow

\frac{2 b}{\sin b} - e^{b \cot b} \ln 2 = 0 [approximate]

t = - b (\cot b - i)

x = e^{b \cot b} (\cos b - i \sin b)

2 solutions \notin R :

b \approx \pm 7.45408757255

x \approx 9.09072986075 \pm 21 . 5079503505 i

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