Exponent Equation Common Factor

[Dalmain]

Good day All

Hope you are keeping well

So I understand up to the point where the common factor is e^-1e^x, surely the common factor would be e^-1e^2. I don't understand how the e^x is part of the common factor. Any help is always appreciated.

Thank you in advance

 

[lex]

An unbelievable solution.

[MATH]\begin{align*} & \;e^{x+1}+e^{x-1}\\ = & \;e^{2+x-1}+e^{x-1}\\ = & \;e^{2}e^{x-1}+e^{x-1}\\ = & \;(e^{2}+1)e^{x-1}\\ \end{align*}[/MATH]
If you are particularly interested, looking at their nonsense (but I wouldn't pay any attention to it):

is

[MATH]e^{x}e^{-1}e^{2}+e^{x}e^{-1}[/MATH]

 

[Dalmain]

Hi Lex

Thank you for your reply, I don't mean to seem slow but I'm still not sure how the first term of e^x+1 becomes e^2+x-1?

 

[lex]

Don't worry.
I've just written x+1 as x-1+2
So [MATH]e^{x+1}[/MATH] can be written as [MATH]e^{x-1+2}[/MATH]which is [MATH]e^{(x-1)+2}[/MATH]which is [MATH]e^{(x-1)}.e^2 \hspace2ex[/MATH] using the rule [MATH]a^{b+c}=a^b.a^c[/MATH]
(The idea being to manufacture a [MATH]e^{x-1}[/MATH] in the first term, which I can then factor out of both terms).

 

[JeffM]

It is odd though correct

[MATH]x + 1 = 1 + x = 2 - 1 + x = 2 + (-1 + x) \implies e^{x+1} = e^{2} * e^{-1+x}.[/MATH]
[MATH]x - 1 = -1 + x = 1(-1 + x) \implies e^{x-1} = 1 * e^{-1+x}.[/MATH]
[MATH]\therefore e^{x+1} + e^{x-1} = (e^2 + 1)e^{-1 + x}.[/MATH]
Of course what is sane is

[MATH]e^{x+1} + e^{x-1} = e^x \left ( e + e^{-1} \right ) = e^x \left (e + \dfrac{1}{e} \right ) = e^x * \dfrac{e^2 + 1}{e}.[/MATH]
Notice that

[MATH](e^2 + 1)e^{-1 +x} = (e^2 + 1) * e^{-1} * e^x = e^x * (e^2 + 1) * \dfrac{1}{e} = e^x * \dfrac{e^2 + 1}{e}.[/MATH]

 

[Dalmain]

Hi Lex and JeffM

Thank you both very much for your feedback, I see what has been done now. It does seem like a weird way to it, but everyday is a school day. Thank you again