• Oct 8th 2009, 03:27 PM
theredqueentheory
Why does e^x - e^x=e^x?
and why does e^x + e^x = e^x?

This doesn't make sense. I know that the derivative of e^x=e^x, that makes sense. But the adding and subtracting 2 of them to = the same thing is weird.

There are several examples in my book that show this. Please explain?
• Oct 8th 2009, 03:31 PM
artvandalay11
Quote:

Originally Posted by theredqueentheory
Why does e^x - e^x=e^x?
and why does e^x + e^x = e^x?

This doesn't make sense. I know that the derivative of e^x=e^x, that makes sense. But the adding and subtracting 2 of them to = the same thing is weird.

There are several examples in my book that show this. Please explain?

I would highly advise looking in your book one more time, because what you've written simply isn't true,

y-y=0 for any and all y, including y=e^x

y+y=2y for any and all y, including y=e^x
• Oct 8th 2009, 03:36 PM
theredqueentheory
thanks but....
I could be wrong but could several other websites be wrong? I suppose they could...please see this about the problem:

Derivatives of exponential and logarithmic functions - An approach to calculus

"The derivative of ex with respect to x
is equal to ex."
• Oct 8th 2009, 03:42 PM
artvandalay11
I am fully aware that the derivative of $e^x=e^x$
and that's what makes it special, but you wrote

$e^x+e^x=e^x$

and

$e^x-e^x=e^x$ which is nonsense
• Oct 8th 2009, 03:58 PM
haebin
derivative of e^x = e^x

and i looked whole pages you stated

e^x - e^x = e^x
e^x + e^x = e^x

which doesn't make sense to me...?

since e^1 = 2.718281828
and in (e^x - e^x) here,
x = x
so
e^x - e^x = 0

e^x + e^x = 2e^x i guess
• Oct 8th 2009, 04:14 PM
theredqueentheory
yes, nonsense.
I agree, it is nonsense. Which is why I don't understand it.
For example, a problem has (2+3)e^-3x + (2+3)e^-3x = 10^e-3x.

Only one ^e-3x. So why when they add two of them does it only equal one?
• Oct 8th 2009, 04:18 PM
mr fantastic
Quote:

Originally Posted by theredqueentheory
I agree, it is nonsense. Which is why I don't understand it.
For example, a problem has (2+3)e^-3x + (2+3)e^-3x = 10^e-3x.

Only one ^e-3x. So why when they add two of them does it only equal one?

2 + 3 = 5. So you have (5 + 5) = 10 lots of e^{-3x}.
• Oct 8th 2009, 06:29 PM
theredqueentheory
um...
so are you saying that e^{-3x} is used as a suffix in this case, and even though the 5 and 5 are added together (thanks for helping with that sum LOL) the e^{-3x} is not added?(Worried)
• Oct 8th 2009, 07:21 PM
mr fantastic
Quote:

Originally Posted by theredqueentheory
so are you saying that e^{-3x} is used as a suffix in this case, and even though the 5 and 5 are added together (thanks for helping with that sum LOL) the e^{-3x} is not added?(Worried)

Let A represent e^{-3x}. It's very obvious that 5A + 5A = 10A.
• Oct 8th 2009, 07:52 PM
theredqueentheory
ok...
It may be obvious to super smart folks like you, I'm just a lowly biotechnologist and my current job in cancer research doesn't require calculus so I am resigned to being referred to as dumb. :) Condescend away!!! I don't mind at all.
Thank you for elucidating that you do not add the suffix together. Then why was everyone in the posts above telling me that if you add two of them together it does not equal one?
• Oct 8th 2009, 08:01 PM
artvandalay11
Quote:

Originally Posted by theredqueentheory
It may be obvious to super smart folks like you, I'm just a lowly biotechnologist and my current job in cancer research doesn't require calculus so I am resigned to being referred to as dumb. :) Condescend away!!! I don't mind at all.
Thank you for elucidating that you do not add the suffix together. Then why was everyone in the posts above telling me that if you add two of them together it does not equal one?

Not trying to be condescending here, but here's by best shot

one "e to the x" plus one "e to the x" equals 2 "e to the x"

You add the coefficients and keep $e^x$ the same

This is not any different from adding
$

3x^2
$
to $5x^2$

You have 3 of "something" and you add 5 of "something" so you get 8 "somethings" i.e. $8x^2$

now read what you originally wrote

$e^x+e^x=e^x$ FALSE
• Oct 8th 2009, 08:12 PM
theredqueentheory
Thank YOU!
I am so glad that math geniuses like you all exist. It keeps the rest of us from going insane, I tell you, INSANE! :) Thank you tons, I completely understand now. (Clapping)
I give you the imaginary medal of "Smarty Pants & Extra Patient Math Helper and Super Explainer of Concepts".
(sorry, it's late and I've done about 12 hours of calculus today, seriously)
• Oct 9th 2009, 07:02 AM
HallsofIvy
I learned that ac+ bc= (a+ b)c in the seventh grade. And that's all that is being used here: 3e^x+ 5e^x= (3+ 5)e^x= 8e^x. I will concede that I did not start calling that the "distributive law" until much later.