What is (2e^{2t}(2t-1)-4e^t)e^{-2t} ?

The solution to (2e^{2t}(2t-1)-4e^t)e^{-2t} is (2(e^t(2t-1)-2))/(e^t)

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simplify (2e^{2t}(2t-1)-4e^t)e^{-2t}

simplify

(2 e^{2 t} (2 t - 1) - 4 e^{t}) e^{- 2 t}

\frac{2 ( e ^{t} ( 2 t - 1 ) - 2 )}{e ^{t}}

Solution steps

(2 e^{2 t} (2 t - 1) - 4 e^{t}) e^{- 2 t}

Apply exponent rule:

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

= \frac{1}{e ^{2 t}} (- 4 e^{t} + 2 e^{2 t} (2 t - 1))

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot ( 2 e ^{2 t} ( 2 t - 1 ) - 4 e ^{t} )}{e ^{2 t}}

1 \cdot (2 e^{2 t} (2 t - 1) - 4 e^{t}) = 2 e^{2 t} (2 t - 1) - 4 e^{t}

= \frac{2 e ^{2 t} ( 2 t - 1 ) - 4 e ^{t}}{e ^{2 t}}

Factor

2 e^{2 t} (2 t - 1) - 4 e^{t} : 2 e^{t} (e^{t} (- 1 + 2 t) - 2)

= \frac{2 e ^{t} ( e ^{t} ( - 1 + 2 t ) - 2 )}{e ^{2 t}}

Cancel

\frac{2 e ^{t} ( e ^{t} ( - 1 + 2 t ) - 2 )}{e ^{2 t}} : \frac{2 ( e ^{t} ( - 1 + 2 t ) - 2 )}{e ^{t}}

= \frac{2 ( e ^{t} ( - 1 + 2 t ) - 2 )}{e ^{t}}

= \frac{2 ( e ^{t} ( 2 t - 1 ) - 2 )}{e ^{t}}

\frac{x + 1}{x + 1 - 1}

e^{- 0.02 \cdot 50} (\frac{( 0.02 \cdot 50 ) ^{5}}{5 !})

(1.26 \cdot 1 0^{- 11}) (7.94 \cdot 1 0^{- 4})

\frac{53}{200} \cdot e^{(- \frac{200}{0.17} \cdot 4.3 \cdot 1 0^{- 4})}

2 e^{- 0.03 (\frac{0.5}{12})}

What is (2e^{2t}(2t-1)-4e^t)e^{-2t} ?
The solution to (2e^{2t}(2t-1)-4e^t)e^{-2t} is (2(e^t(2t-1)-2))/(e^t)