ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある微分積分 >

solvefor y,ln(xy9)=xy

解

解く

y, ln (x y 9) = x y

解

y = - \frac{W _{- 1} ( - \frac{1}{9} )}{x} {x < 0 or x > 0}, y = - \frac{W _{0} ( - \frac{1}{9} )}{x} {x < 0 or x > 0}

解答ステップ

ln (x y \cdot 9) = x y

f (x) = g (x)

ならば,

a^{f (x)} = a^{g (x)}

e^{l n (x y \cdot 9)} = e^{x y}

簡素化

e^{l n (x y \cdot 9)} : 9 y x

9 y x = e^{x y}

解く

9 y x = e^{x y} : y = - \frac{W _{- 1} ( - \frac{1}{9} )}{x}, y = - \frac{W _{0} ( - \frac{1}{9} )}{x}

y = - \frac{W _{- 1} ( - \frac{1}{9} )}{x}, y = - \frac{W _{0} ( - \frac{1}{9} )}{x}

解を検算する:

y = - \frac{W _{- 1} ( - \frac{1}{9} )}{x} {x < 0 or x > 0}, y = - \frac{W _{0} ( - \frac{1}{9} )}{x} {x < 0 or x > 0}

解答は

y = - \frac{W _{- 1} ( - \frac{1}{9} )}{x} {x < 0 or x > 0}, y = - \frac{W _{0} ( - \frac{1}{9} )}{x} {x < 0 or x > 0}

グラフ

人気の例

面積 y= 1/2 x^3+2,y=x+1,x=0,x=2

a re a y = \frac{1}{2} x^{3} + 2, y = x + 1, x = 0, x = 2

limit as x approaches infinity of 1+(2^{x+1})/(2^x-1)

x \to \infty lim (1 + \frac{2 ^{x + 1}}{2 ^{x} - 1})

integral of 3/(ysqrt(4-(ln(y))^2))

\int \frac{3}{y 4 - ( ln ( y ) ) ^{2}} d y

sum from n=1 to infinity of 1/((n+7)n)

n = 1 \sum \infty \frac{1}{( n + 7 ) n}

tangent f(x)=sqrt(x-2),(6,2)

t an g e n t f (x) = x - 2, (6, 2)