zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的代数 >

4^x-2=2^6x

解答

4^{x} - 2 = 2^{6} x

解答

x = - \frac{16 W _{- 1} ( - \frac{l n ( 2 )}{32 e ^{\frac{l n ( 2 )}{16}}} ) + ln ( 2 )}{32 ln ( 2 )}, x = - \frac{16 W _{0} ( - \frac{l n ( 2 )}{32 e ^{\frac{l n ( 2 )}{16}}} ) + ln ( 2 )}{32 ln ( 2 )}

+1

十进制

x = 4.00683 \dots, x = - 0.01596 \dots

求解步骤

4^{x} - 2 = 2^{6} x

将

4^{x} - 2 = 2^{6} x

调整为朗伯形式:

1 = e^{- 2 l n (2) x} (64 x + 2)

用

2 ln (2) (- x - \frac{1}{32}) = u

和

x = - \frac{16 u + ln ( 2 )}{32 ln ( 2 )}

改写方程式

1 = e^{- 2 l n (2) (- \frac{16 u + l n ( 2 )}{32 l n ( 2 )})} (64 (- \frac{16 u + ln ( 2 )}{32 ln ( 2 )}) + 2)

化简

1 = e^{\frac{16 u + l n ( 2 )}{16}} (- \frac{2 ( 16 u + ln ( 2 ) )}{ln ( 2 )} + 2)

改写

1 = e^{\frac{16 u + l n ( 2 )}{16}} (- \frac{2 ( 16 u + ln ( 2 ) )}{ln ( 2 )} + 2)

为朗伯形式:

\frac{e ^{\frac{16 u + l n ( 2 )}{16}} u}{e ^{\frac{l n ( 2 )}{16}}} = - \frac{ln ( 2 )}{32 e ^{\frac{l n ( 2 )}{16}}}

解

\frac{e ^{\frac{16 u + l n ( 2 )}{16}} u}{e ^{\frac{l n ( 2 )}{16}}} = - \frac{ln ( 2 )}{32 e ^{\frac{l n ( 2 )}{16}}} : u = W_{- 1} (- \frac{ln ( 2 )}{32 e ^{\frac{l n ( 2 )}{16}}}), u = W_{0} (- \frac{ln ( 2 )}{32 e ^{\frac{l n ( 2 )}{16}}})

u = W_{- 1} (- \frac{ln ( 2 )}{32 e ^{\frac{l n ( 2 )}{16}}}), u = W_{0} (- \frac{ln ( 2 )}{32 e ^{\frac{l n ( 2 )}{16}}})

代回

u = 2 ln (2) (- x - \frac{1}{32})

，求解

x

解

2 ln (2) (- x - \frac{1}{32}) = W_{- 1} (- \frac{ln ( 2 )}{32 e ^{\frac{l n ( 2 )}{16}}}) : x = - \frac{16 W _{- 1} ( - \frac{l n ( 2 )}{32 e ^{\frac{l n ( 2 )}{16}}} ) + ln ( 2 )}{32 ln ( 2 )}

解

2 ln (2) (- x - \frac{1}{32}) = W_{0} (- \frac{ln ( 2 )}{32 e ^{\frac{l n ( 2 )}{16}}}) : x = - \frac{16 W _{0} ( - \frac{l n ( 2 )}{32 e ^{\frac{l n ( 2 )}{16}}} ) + ln ( 2 )}{32 ln ( 2 )}

x = - \frac{16 W _{- 1} ( - \frac{l n ( 2 )}{32 e ^{\frac{l n ( 2 )}{16}}} ) + ln ( 2 )}{32 ln ( 2 )}, x = - \frac{16 W _{0} ( - \frac{l n ( 2 )}{32 e ^{\frac{l n ( 2 )}{16}}} ) + ln ( 2 )}{32 ln ( 2 )}

作图

流行的例子

3^{x-1}= 1/(2^{x+5)}

3^{x - 1} = \frac{1}{2 ^{x + 5}}

2^{(x+2)(x^2-5x+6)}=1

2^{(x + 2) (x^{2} - 5 x + 6)} = 1

solvefor n,a=p(1+i)^n

so l v e f or n, a = p (1 + i)^{n}

3^{2 y + 1} = 7^{y}

((2^x+1))/((2^x-1))=5

\frac{( 2 ^{x} + 1 )}{( 2 ^{x} - 1 )} = 5