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受欢迎的代数 >

solvefor a,x(a+b)=x^{ab}

解答

求解

a, x (a + b) = x^{ab}

解答

a = - \frac{W ( - \frac{b l n ( x )}{x ^{b^{2} + 1}} ) + b ^{2} ln ( x )}{b ln ( x )}

求解步骤

x (a + b) = x^{ab}

将

x (a + b) = x^{ab}

调整为朗伯形式:

e^{l n (x) (1 - ab)} (a + b) = 1

用

(- ab - b^{2}) ln (x) = u

和

a = - \frac{u + b ^{2} ln ( x )}{b ln ( x )}

改写方程式

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

化简

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

改写

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

为朗伯形式:

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

解

e^{u} u = - \frac{b ln ( x )}{x ^{b^{2} + 1}} : u = W (- \frac{b ln ( x )}{x ^{b^{2} + 1}})

u = W (- \frac{b ln ( x )}{x ^{b^{2} + 1}})

代回

u = (- ab - b^{2}) ln (x)

，求解

a

解

(- ab - b^{2}) ln (x) = W (- \frac{b ln ( x )}{x ^{b^{2} + 1}}) : a = - \frac{W ( - \frac{b l n ( x )}{x ^{b^{2} + 1}} ) + b ^{2} ln ( x )}{b ln ( x )}

a = - \frac{W ( - \frac{b l n ( x )}{x ^{b^{2} + 1}} ) + b ^{2} ln ( x )}{b ln ( x )}

作图

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