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Popular >

m^{sqrt(m)}= 1/2

Solución

m^{m} = \frac{1}{2}

Solución

m = \frac{1}{16}, m = \frac{1}{4}

+1

Notación decimal

m = 0.0625, m = 0.25

Pasos de solución

m^{m} = \frac{1}{2}

Preparar

m^{m} = \frac{1}{2}

para la forma de Lambert:

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

Re-escribir la ecuación con

- \frac{ln ( 2 )}{m} = u

y

m = \frac{ln ^{2} ( 2 )}{u ^{2}}

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

Resolver

(\frac{ln ^{2} ( 2 )}{u ^{2}}) e^{- u} = 1 : u = - 4 ln (2), u = - 2 ln (2), u = 2 W_{0} (\frac{ln ( 2 )}{2})

u = - 4 ln (2), u = - 2 ln (2), u = 2 W_{0} (\frac{ln ( 2 )}{2})

Sustituir hacia atrás la

u = - \frac{ln ( 2 )}{m},

resolver para

m

Resolver

- \frac{ln ( 2 )}{m} = - 4 ln (2) : m = \frac{1}{16}

Resolver

- \frac{ln ( 2 )}{m} = - 2 ln (2) : m = \frac{1}{4}

Resolver

- \frac{ln ( 2 )}{m} = 2 W_{0} (\frac{ln ( 2 )}{2}) :

Sin solución para

m \in R

m = \frac{1}{16}, m = \frac{1}{4}

Verificar las soluciones:

m = \frac{1}{16}

Verdadero

, m = \frac{1}{4}

Verdadero

Las soluciones son

m = \frac{1}{16}, m = \frac{1}{4}

Verificar las soluciones:

m = \frac{1}{16}

Verdadero

, m = \frac{1}{4}

Verdadero

Las soluciones son

m = \frac{1}{16}, m = \frac{1}{4}

Ejemplos populares

e^{x} \cdot x - e^{x} = 0

e^{- x^{2}} - 0.5 = 0

10=58*e^{-90k}+2

10 = 58 \cdot e^{- 90 k} + 2

x ln (e^{x}) + x = 12

x^{10} = 2^{x}