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人気のある代数 >

2^x-5x+2=0

解

2^{x} - 5 x + 2 = 0

解

x = - \frac{5 W _{- 1} ( - \frac{e ^{\frac{2 l n ( 2 )}{5}} l n ( 2 )}{5} ) - 2 ln ( 2 )}{5 ln ( 2 )}, x = - \frac{5 W _{0} ( - \frac{e ^{\frac{2 l n ( 2 )}{5}} l n ( 2 )}{5} ) - 2 ln ( 2 )}{5 ln ( 2 )}

+1

十進法表記

x = 4.27679 \dots, x = 0.73224 \dots

解答ステップ

2^{x} - 5 x + 2 = 0

ランベルト形式向けに

2^{x} - 5 x + 2 = 0

を準備する:

1 = - e^{- l n (2) x} (- 5 x + 2)

equationを

(- x + \frac{2}{5}) ln (2) = u

と以下で書き換える:

x = - \frac{5 u - 2 ln ( 2 )}{5 ln ( 2 )}

1 = - e^{- l n (2) (- \frac{5 u - 2 l n ( 2 )}{5 l n ( 2 )})} (- 5 (- \frac{5 u - 2 ln ( 2 )}{5 ln ( 2 )}) + 2)

簡素化

1 = - e^{\frac{5 u - 2 l n ( 2 )}{5}} (\frac{5 u - 2 ln ( 2 )}{ln ( 2 )} + 2)

ランベルト形式で

1 = - e^{\frac{5 u - 2 l n ( 2 )}{5}} (\frac{5 u - 2 ln ( 2 )}{ln ( 2 )} + 2)

を書き換える:

e^{u} u = - \frac{e ^{\frac{2 l n ( 2 )}{5}} ln ( 2 )}{5}

解く

e^{u} u = - \frac{e ^{\frac{2 l n ( 2 )}{5}} ln ( 2 )}{5} : u = W_{- 1} (- \frac{e ^{\frac{2 l n ( 2 )}{5}} ln ( 2 )}{5}), u = W_{0} (- \frac{e ^{\frac{2 l n ( 2 )}{5}} ln ( 2 )}{5})

u = W_{- 1} (- \frac{e ^{\frac{2 l n ( 2 )}{5}} ln ( 2 )}{5}), u = W_{0} (- \frac{e ^{\frac{2 l n ( 2 )}{5}} ln ( 2 )}{5})

再び

u = (- x + \frac{2}{5}) ln (2)

に置き換えて以下を解く:

x

解く

(- x + \frac{2}{5}) ln (2) = W_{- 1} (- \frac{e ^{\frac{2 l n ( 2 )}{5}} ln ( 2 )}{5}) : x = - \frac{5 W _{- 1} ( - \frac{e ^{\frac{2 l n ( 2 )}{5}} l n ( 2 )}{5} ) - 2 ln ( 2 )}{5 ln ( 2 )}

解く

(- x + \frac{2}{5}) ln (2) = W_{0} (- \frac{e ^{\frac{2 l n ( 2 )}{5}} ln ( 2 )}{5}) : x = - \frac{5 W _{0} ( - \frac{e ^{\frac{2 l n ( 2 )}{5}} l n ( 2 )}{5} ) - 2 ln ( 2 )}{5 ln ( 2 )}

x = - \frac{5 W _{- 1} ( - \frac{e ^{\frac{2 l n ( 2 )}{5}} l n ( 2 )}{5} ) - 2 ln ( 2 )}{5 ln ( 2 )}, x = - \frac{5 W _{0} ( - \frac{e ^{\frac{2 l n ( 2 )}{5}} l n ( 2 )}{5} ) - 2 ln ( 2 )}{5 ln ( 2 )}

グラフ

人気の例

3^{- x} = 2

15 + 4^{x} = 31

2^{2x}-3*2^{x+2}+32=0

2^{2 x} - 3 \cdot 2^{x + 2} + 32 = 0

1 \cdot 1 0^{x} = 3

4^{x} + 2^{x} = 8