What is e^{L*(e^t-1)}*Le^{-Lx} ?

The solution to e^{L*(e^t-1)}*Le^{-Lx} is Le^{-Lx+L(e^t-1)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify e^{L(e^t-1)}Le^{-Lx}

simplify

e^{L \cdot (e^{t} - 1)} \cdot L e^{- Lx}

L e^{- Lx + L (e^{t} - 1)}

Solution steps

e^{L (e^{t} - 1)} L e^{- Lx}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

e^{L (e^{t} - 1)} e^{- Lx} = e^{L (e^{t} - 1) - Lx}

= L e^{L (e^{t} - 1) - Lx}

- 3.3 e^{- 0.3184}

4905 (7.854 \cdot 1 0^{- 3})

m n

0.53 \cdot 1 0^{- 10} \cdot 16 \cdot 2 π

What is e^{L*(e^t-1)}*Le^{-Lx} ?
The solution to e^{L*(e^t-1)}*Le^{-Lx} is Le^{-Lx+L(e^t-1)}