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Let $a: (0,\infty)\rightarrow (0,\infty)$ and let $a_s\equiv a(s)$ $\forall s>0$.

Let $f: \mathbb{R}\rightarrow (0,\infty)$ where $f(x)\equiv \log(a_{e^x})$.

Could you help me to show that $f(x)$ satisfies the Cauchy Function Equation $$ f(x+y)=f(x)+f(y) $$ (e.g., see the end of p.3 here)

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    $\begingroup$ What does $a_{e^{x}}$ stand for? $\endgroup$ – Kabo Murphy Apr 13 '18 at 11:55
  • $\begingroup$ @KaviRamaMurthy thanks, see the edit in my question $\endgroup$ – STF Apr 17 '18 at 10:02
  • $\begingroup$ Surely false if $a$ is arbitrary. $\endgroup$ – Kabo Murphy Apr 17 '18 at 10:17
  • $\begingroup$ @KaviRamaMurthy Thank you. I realised that the notes I linked are referred to a specific $a$. $\endgroup$ – STF Apr 17 '18 at 10:36
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If $a(x)=x+1$ then the equation $f(x+y)=f(x)+f(y)$ does not hold for any $x,y>0$

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I suppose that you are asking about:

$$f(x)=\log_a{e^x}$$

$$f(x+y)=\log_a{e^{x+y}} = \log_a{(e^x e^y)} = \log_a{e^x} + \log_a{e^y} = f(x)+f(y)$$

Done.

Edit: But that was all just unnecessary "gymnastics".

$$f(x)=\log_a{e^x}=\frac{\ln e^x}{\ln a}=\frac{x}{\ln a}$$

The rest is trivial.

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  • $\begingroup$ I have clarified my question $\endgroup$ – STF Apr 17 '18 at 10:01

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