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人気のある代数 >

3^x+2x-5=0

解

3^{x} + 2 x - 5 = 0

解

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

+1

十進法表記

x = 1

解答ステップ

3^{x} + 2 x - 5 = 0

ランベルト形式向けに

3^{x} + 2 x - 5 = 0

を準備する:

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

equationを

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

と以下で書き換える:

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

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

簡素化

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

ランベルト形式で

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

を書き換える:

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

解く

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

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

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

グラフ

人気の例

2^{5 x + 2} = 8^{9}

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

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

solvefor y,((2^y-1))/((2^y+1))=x

so l v e f or y, \frac{( 2 ^{y} - 1 )}{( 2 ^{y} + 1 )} = x

3^{x} = 171

2^{1+x}+2^{1-x}=5

2^{1 + x} + 2^{1 - x} = 5