ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある代数 >

2x-4=3^{-x}+1

解

2 x - 4 = 3^{- x} + 1

解

x = \frac{W _{0} ( \frac{l n ( 3 )}{2 e ^{\frac{5 l n ( 3 )}{2}}} )}{ln ( 3 )} + \frac{5}{2}

+1

十進法表記

x = 2.53100 \dots

解答ステップ

2 x - 4 = 3^{- x} + 1

ランベルト形式向けに

2 x - 4 = 3^{- x} + 1

を準備する:

1 = (2 x - 5) e^{l n (3) x}

equationを

(x - \frac{5}{2}) ln (3) = u

と以下で書き換える:

x = \frac{u}{ln ( 3 )} + \frac{5}{2}

1 = (2 (\frac{u}{ln ( 3 )} + \frac{5}{2}) - 5) e^{l n (3) (\frac{u}{l n ( 3 )} + \frac{5}{2})}

簡素化

1 = \frac{2 e ^{u + \frac{5}{2} l n (3)} u}{ln ( 3 )}

ランベルト形式で

1 = \frac{2 e ^{u + \frac{5}{2} l n (3)} u}{ln ( 3 )}

を書き換える:

\frac{e ^{\frac{2 u + 5 l n ( 3 )}{2}} u}{e ^{\frac{5 l n ( 3 )}{2}}} = \frac{ln ( 3 )}{2 e ^{\frac{5 l n ( 3 )}{2}}}

解く

\frac{e ^{\frac{2 u + 5 l n ( 3 )}{2}} u}{e ^{\frac{5 l n ( 3 )}{2}}} = \frac{ln ( 3 )}{2 e ^{\frac{5 l n ( 3 )}{2}}} : u = W_{0} (\frac{ln ( 3 )}{2 e ^{\frac{5 l n ( 3 )}{2}}})

u = W_{0} (\frac{ln ( 3 )}{2 e ^{\frac{5 l n ( 3 )}{2}}})

再び

u = (x - \frac{5}{2}) ln (3)

に置き換えて以下を解く:

x

解く

(x - \frac{5}{2}) ln (3) = W_{0} (\frac{ln ( 3 )}{2 e ^{\frac{5 l n ( 3 )}{2}}}) : x = \frac{W _{0} ( \frac{l n ( 3 )}{2 e ^{\frac{5 l n ( 3 )}{2}}} )}{ln ( 3 )} + \frac{5}{2}

x = \frac{W _{0} ( \frac{l n ( 3 )}{2 e ^{\frac{5 l n ( 3 )}{2}}} )}{ln ( 3 )} + \frac{5}{2}

グラフ

人気の例

2^{x} - 1 = 2^{6}

(0.04)^{5x-x^2-8}=625

(0.04)^{5 x - x^{2} - 8} = 625

3 = 3^{2 x} - 3^{x} \cdot 2

1 0^{x} = 7000

5 + (3) 4^{x} - 1 = 12