What is the answer to solvefor z,y=e^{-fxz^{2.5}} ?

The answer to solvefor z,y=e^{-fxz^{2.5}} is z=(-(ln(y))/(fx))^{2/5}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solvefor z,y=e^{-fxz^{2.5}}

solve for

z, y = e^{- f x z^{2.5}}

z = (- \frac{ln ( y )}{f x})^{\frac{2}{5}}

Solution steps

y = e^{- f x z^{2.5}}

Apply exponent rules

ln (y) = - f x z^{2.5}

Solve

ln (y) = - f x z^{2.5} : z = (- \frac{ln ( y )}{f x})^{\frac{2}{5}}

z = (- \frac{ln ( y )}{f x})^{\frac{2}{5}}

x, y^{2} = e^{2 x}

93000 \cdot 1.098 7^{x} = 156500

3^{2 x + 1} - 7 \cdot 3^{x} + 4 = 0

e^{x} - e^{- x} = \frac{3}{2}

x, 3 y = - 5^{x} + 7

What is the answer to solvefor z,y=e^{-fxz^{2.5}} ?
The answer to solvefor z,y=e^{-fxz^{2.5}} is z=(-(ln(y))/(fx))^{2/5}