What is e^{-5/2 ln(10-2t)dt} ?

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

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simplify e^{-5/2 ln(10-2t)dt}

simplify

e^{- \frac{5}{2} l n (10 - 2 t) d t}

\frac{1}{( - 2 t + 10 ) ^{\frac{5 t d}{2}}}

Solution steps

e^{- \frac{5}{2} l n (10 - 2 t) d t}

Apply exponent rule:

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

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

e^{\frac{5}{2} l n (10 - 2 t) d t} = (10 - 2 t)^{\frac{5}{2} t d}

= \frac{1}{( 10 - 2 t ) ^{\frac{5}{2} t d}}

(10 - 2 t)^{\frac{5}{2} d t} = (10 - 2 t)^{\frac{5 t d}{2}}

= \frac{1}{( 10 - 2 t ) ^{\frac{5 d t}{2}}}

\frac{d}{d t} ((y^{^{'}} (t))^{2})

(3 x) (6 x^{2})

- 4 a^{4} b^{4}

7 e^{- 93 \cdot 34}

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