Natural logarithm problem

Question

I'm kind of confused on how to solve this problem! Any guidance/advice would be appreciated. Thanks!

$e^{−9}e^{−2}e^{9}$

Answer 1

In general

\begin{align*} a^b a^c = a^{b+c} \end{align*}

Therefore,

\begin{align*} \mathrm{e}^{-9}\mathrm{e}^{-2} \mathrm{e}^{9} & = (\mathrm{e}^{-9}\mathrm{e}^{-2}) \mathrm{e}^{9} \\ & = \mathrm{e}^{-11} \mathrm{e}^{9} \\ & = \mathrm{e}^{-2} \end{align*}

Answer 2

$$e^{−9}e^{−2}e^9=e^{(-9)+(-2)+9}=e^{-9-2+9}=e^{-2}$$

Note that this only works when the exponents have the same base, as is the case here.