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受欢迎的代数 >

e^{x^x}=2

解答

e^{x^{x}} = 2

解答

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

+1

十进制

x = 0.33662 \dots, x = 0.40004 \dots

求解步骤

e^{x^{x}} = 2

将

e^{x^{x}} = 2

调整为朗伯形式:

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

用

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

和

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

改写方程式

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

改写

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

为朗伯形式:

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

解

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

验证解:

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

真

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

真

解为

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

代回

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

，求解

x

解

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

解

\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 ) ) )}

作图

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