# Can I add 2e^(-2x) and 2e^(-x)?

Nov 12, 2017

No.

#### Explanation:

You can’t add things with different variables.
When I’m not sure how something with variables works, I plug in some small numbers to think about how they work.

Here, instead of using $e$, consider ${3}^{2}$ and ${3}^{3}$. How many $3$’s is that?

We don’t know without multiplying out the threes, so we can’t do it with x’s because we don’t know the value of them.

Nov 12, 2017

$= \frac{2 \left(1 + {e}^{x}\right)}{e} ^ \left(2 x\right) = 2 {e}^{- 2 x} \left(1 + {e}^{x}\right)$
as explained below.

#### Explanation:

$2 {e}^{- 2 x} + 2 {e}^{-} x$
$= \left(\frac{2}{e} ^ \left(2 x\right)\right) + \left(\frac{2}{e} ^ x\right)$
As the denominators are different, they can't be simply added.
We have to get LCM, which is nothing but ${e}^{2 x} a s$e^x * e^x = e^(2x)#

$= \frac{2 + 2 {e}^{x}}{e} ^ \left(2 x\right)$

$= \frac{2 \left(1 + {e}^{x}\right)}{e} ^ \left(2 x\right)$

$= 2 {e}^{- 2 x} \left(1 + {e}^{x}\right)$