Compare exp(a(x+y)) to a(exp(x) + exp(y)) [closed]

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

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

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

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).

• Thanks for working with me while I clarified the question. Sorry to confuse everyone else. – MadProgrammer Jun 26 at 15:10

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

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

• “notoriously” seems a bit extra – gen-z ready to perish Jun 25 at 14:26
• @ChaseRyanTaylor: read obviously then. – Yves Daoust Jun 25 at 14:28
• I don’t know what you’re referring to – gen-z ready to perish Jun 25 at 14:29

The expression is not valid. For example if you let $$a=0$$ you get $$1=0$$

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

Careful with typos!

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

i.e.,

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