Question

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

Answer 1

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.

Answer 2

It cannot be added directly.

`= (2(1+e^x))/e^(2x) = 2e^(-2x)(1+e^x)`
as explained below.

Explanation:

`2e^(-2x) + 2e^-x`
`= (2/e^(2x)) + (2/e^x)`
As the denominators are different, they can't be simply added.
We have to get LCM, which is nothing but `e^(2x) as`e^x * e^x = e^(2x)#

`= (2 + 2e^x)/e^(2x)`

`= (2(1+e^x))/e^(2x)`

`= 2e^(-2x)(1+e^x)`