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

2 Answers
Nov 12, 2017

Answer:

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

Answer:

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)#