# How do you simplify  e^(-3) e^(-7) e^ (-4)?

${e}^{-} 3 {e}^{-} 7 {e}^{-} 4 = {e}^{-} 14$
Since ${x}^{a} {x}^{b} = {x}^{a + b}$, then ${e}^{-} 3 {e}^{-} 7 = {e}^{- 3 + \left(- 7\right)} = {e}^{-} 10$
Further ${e}^{-} 3 {e}^{-} 7 {e}^{-} 4 = {e}^{-} 10 {e}^{-} 4 = {e}^{- 10 + \left(- 4\right)} = {e}^{-} 14$
Thus, ${e}^{-} 3 {e}^{-} 7 {e}^{-} 4 = {e}^{-} 14$