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人気のある代数 >

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

解

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

解

x = - \frac{8 W _{- 1} ( - \frac{e ^{\frac{5 l n ( 2 )}{8}} l n ( 2 )}{8} ) - 5 ln ( 2 )}{8 ln ( 2 )}, x = - \frac{8 W _{0} ( - \frac{e ^{\frac{5 l n ( 2 )}{8}} l n ( 2 )}{8} ) - 5 ln ( 2 )}{8 ln ( 2 )}

+1

十進法表記

x = 5.19089 \dots, x = 0.85037 \dots

解答ステップ

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

ランベルト形式向けに

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

を準備する:

1 = (8 x - 5) e^{- l n (2) x}

equationを

(- x + \frac{5}{8}) ln (2) = u

と以下で書き換える:

x = - \frac{8 u - 5 ln ( 2 )}{8 ln ( 2 )}

1 = (8 (- \frac{8 u - 5 ln ( 2 )}{8 ln ( 2 )}) - 5) e^{- l n (2) (- \frac{8 u - 5 l n ( 2 )}{8 l n ( 2 )})}

簡素化

1 = e^{\frac{8 u - 5 l n ( 2 )}{8}} (- \frac{8 u - 5 ln ( 2 )}{ln ( 2 )} - 5)

ランベルト形式で

1 = e^{\frac{8 u - 5 l n ( 2 )}{8}} (- \frac{8 u - 5 ln ( 2 )}{ln ( 2 )} - 5)

を書き換える:

e^{u} u = - \frac{e ^{\frac{5 l n ( 2 )}{8}} ln ( 2 )}{8}

解く

e^{u} u = - \frac{e ^{\frac{5 l n ( 2 )}{8}} ln ( 2 )}{8} : u = W_{- 1} (- \frac{e ^{\frac{5 l n ( 2 )}{8}} ln ( 2 )}{8}), u = W_{0} (- \frac{e ^{\frac{5 l n ( 2 )}{8}} ln ( 2 )}{8})

u = W_{- 1} (- \frac{e ^{\frac{5 l n ( 2 )}{8}} ln ( 2 )}{8}), u = W_{0} (- \frac{e ^{\frac{5 l n ( 2 )}{8}} ln ( 2 )}{8})

再び

u = (- x + \frac{5}{8}) ln (2)

に置き換えて以下を解く:

x

解く

(- x + \frac{5}{8}) ln (2) = W_{- 1} (- \frac{e ^{\frac{5 l n ( 2 )}{8}} ln ( 2 )}{8}) : x = - \frac{8 W _{- 1} ( - \frac{e ^{\frac{5 l n ( 2 )}{8}} l n ( 2 )}{8} ) - 5 ln ( 2 )}{8 ln ( 2 )}

解く

(- x + \frac{5}{8}) ln (2) = W_{0} (- \frac{e ^{\frac{5 l n ( 2 )}{8}} ln ( 2 )}{8}) : x = - \frac{8 W _{0} ( - \frac{e ^{\frac{5 l n ( 2 )}{8}} l n ( 2 )}{8} ) - 5 ln ( 2 )}{8 ln ( 2 )}

x = - \frac{8 W _{- 1} ( - \frac{e ^{\frac{5 l n ( 2 )}{8}} l n ( 2 )}{8} ) - 5 ln ( 2 )}{8 ln ( 2 )}, x = - \frac{8 W _{0} ( - \frac{e ^{\frac{5 l n ( 2 )}{8}} l n ( 2 )}{8} ) - 5 ln ( 2 )}{8 ln ( 2 )}

グラフ

人気の例

110=120(1-e^{-0.2t})

110 = 120 (1 - e^{- 0.2 t})

log_{10}(0.125^{x-2})=2

lo g_{10} (0.12 5^{x - 2}) = 2

2^{x - 1} = 128

5^{x + 4} = 8

solvefor y,15^{-8y}=4

so l v e f or y, 1 5^{- 8 y} = 4