What is e^{-(5y)/2 ln(10-2x)} ?

The solution to e^{-(5y)/2 ln(10-2x)} is 1/((-2x+10)^{\frac{5y){2}}}

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simplify e^{-(5y)/2 ln(10-2x)}

simplify

e^{- \frac{5 y}{2} l n (10 - 2 x)}

\frac{1}{( - 2 x + 10 ) ^{\frac{5 y}{2}}}

Solution steps

e^{- \frac{5 y}{2} l n (10 - 2 x)}

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \frac{1}{e ^{\frac{5 y}{2} l n (- 2 x + 10)}}

e^{\frac{5 y}{2} l n (10 - 2 x)} = (10 - 2 x)^{\frac{5}{2} y}

= \frac{1}{( 10 - 2 x ) ^{\frac{5}{2} y}}

(10 - 2 x)^{\frac{5}{2} y} = (10 - 2 x)^{\frac{5 y}{2}}

= \frac{1}{( 10 - 2 x ) ^{\frac{5 y}{2}}}

e^{- 3 t} \cdot e^{- j wt}

e^{- 15.75}

0.084 \cdot (7.85 \cdot 1 0^{- 3})

- 6 y^{2} (x + y)^{2}

What is e^{-(5y)/2 ln(10-2x)} ?
The solution to e^{-(5y)/2 ln(10-2x)} is 1/((-2x+10)^{\frac{5y){2}}}