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 >

n=1+(9/10)^n

解

n = 1 + (\frac{9}{10})^{n}

解

n = - \frac{W _{0} ( - \frac{9 l n ( \frac{9}{10} )}{10} ) - ln ( \frac{9}{10} )}{ln ( \frac{9}{10} )}

+1

十進法表記

n = 1.82506 \dots

解答ステップ

n = 1 + (\frac{9}{10})^{n}

ランベルト形式向けに

n = 1 + (\frac{9}{10})^{n}

を準備する:

1 = (n - 1) e^{- l n (\frac{9}{10}) n}

equationを

ln (\frac{9}{10}) (- n + 1) = u

と以下で書き換える:

n = - \frac{u - ln ( \frac{9}{10} )}{ln ( \frac{9}{10} )}

1 = ((- \frac{u - ln ( \frac{9}{10} )}{ln ( \frac{9}{10} )}) - 1) e^{- l n (\frac{9}{10}) (- \frac{u - l n ( \frac{9}{10} )}{l n ( \frac{9}{10} )})}

簡素化

1 = e^{u - l n (\frac{9}{10})} (- \frac{u - ln ( \frac{9}{10} )}{ln ( \frac{9}{10} )} - 1)

ランベルト形式で

1 = e^{u - l n (\frac{9}{10})} (- \frac{u - ln ( \frac{9}{10} )}{ln ( \frac{9}{10} )} - 1)

を書き換える:

e^{u} u = - \frac{9 ln ( \frac{9}{10} )}{10}

解く

e^{u} u = - \frac{9 ln ( \frac{9}{10} )}{10} : u = W_{0} (- \frac{9 ln ( \frac{9}{10} )}{10})

u = W_{0} (- \frac{9 ln ( \frac{9}{10} )}{10})

再び

u = ln (\frac{9}{10}) (- n + 1)

に置き換えて以下を解く:

n

解く

ln (\frac{9}{10}) (- n + 1) = W_{0} (- \frac{9 ln ( \frac{9}{10} )}{10}) : n = - \frac{W _{0} ( - \frac{9 l n ( \frac{9}{10} )}{10} ) - ln ( \frac{9}{10} )}{ln ( \frac{9}{10} )}

n = - \frac{W _{0} ( - \frac{9 l n ( \frac{9}{10} )}{10} ) - ln ( \frac{9}{10} )}{ln ( \frac{9}{10} )}

グラフ

人気の例

2^2x+3+2^x+3=1+2^x

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

(1/32)^{x+2}=8^{x+1}

(\frac{1}{32})^{x + 2} = 8^{x + 1}

2^x-1+2^x+2^x+1=7

2^{x} - 1 + 2^{x} + 2^{x} + 1 = 7

3^{2x-5}+9^{x-2}=4

3^{2 x - 5} + 9^{x - 2} = 4

(1/16)^x=(1/8)^6

(\frac{1}{16})^{x} = (\frac{1}{8})^{6}