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受欢迎的代数 >

(5^x0.8^{x-1})/x =500

解答

\frac{5 ^{x} 0. 8 ^{x - 1}}{x} = 500

解答

x = - \frac{W _{- 1} ( - \frac{l n ( 2 )}{200} )}{2 ln ( 2 )}, x = - \frac{W _{0} ( - \frac{l n ( 2 )}{200} )}{2 ln ( 2 )}

+1

十进制

x = 5.55939 \dots, x = 0.00250 \dots

求解步骤

\frac{5 ^{x} \cdot 0. 8 ^{x - 1}}{x} = 500

将

\frac{5 ^{x} 0. 8 ^{x - 1}}{x} = 500

调整为朗伯形式:

1 = 25 e^{l n (2) (- 2 x + 4)} x

用

- 2 x ln (2) = u

和

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

改写方程式

1 = 25 e^{l n (2) (- 2 (- \frac{u}{2 l n ( 2 )}) + 4)} (- \frac{u}{2 ln ( 2 )})

化简

1 = - \frac{25 e ^{u + 4 l n (2)} u}{2 ln ( 2 )}

改写

1 = - \frac{25 e ^{u + 4 l n (2)} u}{2 ln ( 2 )}

为朗伯形式:

e^{u} u = - \frac{ln ( 2 )}{200}

解

e^{u} u = - \frac{ln ( 2 )}{200} : u = W_{- 1} (- \frac{ln ( 2 )}{200}), u = W_{0} (- \frac{ln ( 2 )}{200})

u = W_{- 1} (- \frac{ln ( 2 )}{200}), u = W_{0} (- \frac{ln ( 2 )}{200})

代回

u = - 2 x ln (2)

，求解

x

解

- 2 x ln (2) = W_{- 1} (- \frac{ln ( 2 )}{200}) : x = - \frac{W _{- 1} ( - \frac{l n ( 2 )}{200} )}{2 ln ( 2 )}

解

- 2 x ln (2) = W_{0} (- \frac{ln ( 2 )}{200}) : x = - \frac{W _{0} ( - \frac{l n ( 2 )}{200} )}{2 ln ( 2 )}

x = - \frac{W _{- 1} ( - \frac{l n ( 2 )}{200} )}{2 ln ( 2 )}, x = - \frac{W _{0} ( - \frac{l n ( 2 )}{200} )}{2 ln ( 2 )}

验证解:

x = - \frac{W _{- 1} ( - \frac{l n ( 2 )}{200} )}{2 ln ( 2 )}

真

, x = - \frac{W _{0} ( - \frac{l n ( 2 )}{200} )}{2 ln ( 2 )}

真

解为

x = - \frac{W _{- 1} ( - \frac{l n ( 2 )}{200} )}{2 ln ( 2 )}, x = - \frac{W _{0} ( - \frac{l n ( 2 )}{200} )}{2 ln ( 2 )}

验证解:

x = - \frac{W _{- 1} ( - \frac{l n ( 2 )}{200} )}{2 ln ( 2 )}

真

, x = - \frac{W _{0} ( - \frac{l n ( 2 )}{200} )}{2 ln ( 2 )}

真

解为

x = - \frac{W _{- 1} ( - \frac{l n ( 2 )}{200} )}{2 ln ( 2 )}, x = - \frac{W _{0} ( - \frac{l n ( 2 )}{200} )}{2 ln ( 2 )}

作图

流行的例子

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

(k + 1) 2^{k} = 32

1/(2^{x-4)}-1/(2^{x-2)}=3

\frac{1}{2 ^{x - 4}} - \frac{1}{2 ^{x - 2}} = 3

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

3^{x} = 5^{x - 2}