Can anyone please explain how $$\exp[y/\alpha]=\exp[y-\log\alpha]$$ ?

I tried as :


I think $$\exp(y)\exp[-\log\alpha],$$ can be written as $$\exp[y-\log\alpha].$$

But not $$\exp(y/\exp[\log\alpha]),$$ can be written as $$\exp[y-\log\alpha].$$


The left-hand side should read $\exp(y)/a$
Can you take it from there?
$$\exp(y)/a=\exp(y)/\exp(\log a)=\exp(y-\log a)$$ On the other hand, $$\exp(6/3)=\exp(2)=e^2=7.389\\ \exp(6)/3=(e^6)/3=403.429/3=134.476$$

  • $\begingroup$ Is $\exp(y/a)=(1/a)\exp(y)$? $\endgroup$ – ABC Sep 28 '15 at 6:43
  • $\begingroup$ No, $exp(y/a)=\sqrt[a]{\exp(y)}$ is quite different. $\endgroup$ – Empy2 Sep 28 '15 at 6:46
  • $\begingroup$ Could you please add a little more explanation to your answer ? $\endgroup$ – ABC Sep 28 '15 at 6:49

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