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I would like to compare $\exp(a(x+y))$ to $a(\exp(x) + \exp(y))$ for $a>0$. How do I approach this?

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closed as off-topic by lisyarus, Adrian Keister, kingW3, The Count, Lord Shark the Unknown Jun 26 at 3:58

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  • $\begingroup$ This is false as stated. For the related true statement, please tell us what you are assuming. $\endgroup$ – Dunham Jun 25 at 13:55
  • $\begingroup$ Ok, please tell me why. $\endgroup$ – MadProgrammer Jun 25 at 13:56
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    $\begingroup$ take x=y=0 and a=1. $\endgroup$ – Dunham Jun 25 at 13:57
  • $\begingroup$ I'm working on this with a constant multiplied out front: math.stackexchange.com/questions/285227/… $\endgroup$ – MadProgrammer Jun 25 at 13:59
  • $\begingroup$ Please look again at that post. The right hand side is a multiplication, not addition. $\endgroup$ – mjw Jun 25 at 14:01
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It is probably easiest to restrict to lines in the plane and consider what happens. Here are a few cases:

Case 1: $y = 0$ (By symmetry, similar to $x=0$)

We are comparing $\exp(ax)$ to $a\exp(x)$.

Taking the natural log of each, we see that they are equal when $x = \log(a)/(a-1)$.

If $a>1$, $\exp(ax)$ grows faster as $x\rightarrow \infty$, and decays to 0 faster as $x\rightarrow -\infty$

If $a<1$, $a\exp(x)$ grows faster as $x\rightarrow \infty$, and decays to 0 faster as $x\rightarrow -\infty$

Case 2: $x = y$

We are comparing $\exp(2ax)$ to $2a\exp(x)$.

The analysis is very similar, with $a$ replaced by $2a$.

Note

Without too much work, you can also do asymptotics for $y = mx$ (arbitrary line of slope $m$ in the plane).

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  • $\begingroup$ Thanks for working with me while I clarified the question. Sorry to confuse everyone else. $\endgroup$ – MadProgrammer Jun 26 at 15:10
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Set $y=0$ and your identity becomes

$$\exp(ax)=a\exp(x),$$ which is notoriously false.

Still not convinced ? With $x=1$,

$$\exp(a)=a\,e\ ???$$


A correct statement is

$$(\exp(x+y))^a=\exp(a(x+y))=\exp(ax+ay)=\\\exp(ax)\exp(ay)=(\exp(x))^a(\exp(y))^a=(\exp(x)\exp(y))^a.$$

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  • $\begingroup$ “notoriously” seems a bit extra $\endgroup$ – gen-z ready to perish Jun 25 at 14:26
  • $\begingroup$ @ChaseRyanTaylor: read obviously then. $\endgroup$ – Yves Daoust Jun 25 at 14:28
  • $\begingroup$ I don’t know what you’re referring to $\endgroup$ – gen-z ready to perish Jun 25 at 14:29
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The expression is not valid. For example if you let $a=0$ you get $1=0$

What you may ask is $$e^{a( x+y)}=e^{ax}e^{ay}$$

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Careful with typos!

$$e^{a(x+y)}=e^{ax} e^{a y}$$

i.e.,

$$\exp(a(x+y))=\exp(ax)\exp(a y)$$

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