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人気のある代数 >

e^{x^x}=2

解

e^{x^{x}} = 2

解

x = \frac{ln ( ln ( 2 ) )}{W _{- 1} ( ln ( ln ( 2 ) ) )}, x = \frac{ln ( ln ( 2 ) )}{W _{0} ( ln ( ln ( 2 ) ) )}

+1

十進法表記

x = 0.33662 \dots, x = 0.40004 \dots

解答ステップ

e^{x^{x}} = 2

ランベルト形式向けに

e^{x^{x}} = 2

を準備する:

x e^{- \frac{l n ( l n ( 2 ) )}{x}} = 1

equationを

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

と以下で書き換える:

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

(\frac{ln ( ln ( 2 ) )}{u}) e^{- u} = 1

ランベルト形式で

(\frac{ln ( ln ( 2 ) )}{u}) e^{- u} = 1

を書き換える:

u e^{u} = ln (ln (2))

解く

u e^{u} = ln (ln (2)) : u = W_{- 1} (ln (ln (2))), u = W_{0} (ln (ln (2)))

解を検算する:

u = W_{- 1} (ln (ln (2)))

真

, u = W_{0} (ln (ln (2)))

真

解答は

u = W_{- 1} (ln (ln (2))), u = W_{0} (ln (ln (2)))

再び

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

に置き換えて以下を解く:

x

解く

\frac{ln ( ln ( 2 ) )}{x} = W_{- 1} (ln (ln (2))) : x = \frac{ln ( ln ( 2 ) )}{W _{- 1} ( ln ( ln ( 2 ) ) )}

解く

\frac{ln ( ln ( 2 ) )}{x} = W_{0} (ln (ln (2))) : x = \frac{ln ( ln ( 2 ) )}{W _{0} ( ln ( ln ( 2 ) ) )}

x = \frac{ln ( ln ( 2 ) )}{W _{- 1} ( ln ( ln ( 2 ) ) )}, x = \frac{ln ( ln ( 2 ) )}{W _{0} ( ln ( ln ( 2 ) ) )}

グラフ

人気の例

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