This is a consequence of the exponential rule, but how do I actually prove it to be true?
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$$x>-\infty\implies a^x>a^{-\infty}$$ for real $a>1$ – lab bhattacharjee Apr 09 '14 at 14:26
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3How do you define $e^x$ ? – lhf Apr 09 '14 at 14:27
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$e$ is a positive number. Think about what happens when you multiply a positive number by a positive number. – rookie Apr 09 '14 at 14:39
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what is its inverse ? – derivative Apr 09 '14 at 14:39
3 Answers
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How do you define $e^x$?
The most common way to define it is
$$e^x = \sum_{n = 0}^{\infty} \frac{x^n}{n!}$$
If you take this definition, then if $x > 0$, $e^x$ is the sum of an infinite number of positive terms.
One should show that the series converges, but if you take that for granted, then of course if converges to a positive number.
If $x < 0$ then $e^{-x} > 0$ ; and since $$ e^x \cdot e^{-x} = 1 > 0 \Rightarrow e^x > 0$$ since $e^{-x} > 0$
Ant
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Your approach lacks to study the behavior of $e^{x}$ for $x<0$. Any idea on how to do it using simple considerations ? – jibe Apr 09 '14 at 14:31
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@jibe ops I misread the question! I edited :-) Thanks to both of you for pointing that out :-) – Ant Apr 09 '14 at 14:36
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Every definition of $\exp(x)$ leads to the property, $\exp(a+b)= \exp(a)\exp(b)$. Furthermore every definition also tells us that $\exp$ is defined on the entire real line and takes real values when given a real number as an argument.
So we will consider the exponential function evaluated at $x$ and notice that$x=x/2+x/2$.
$$ \exp(x) = \exp(x/2)\exp(x/2) = (\exp(x/2))^2 > 0 $$
Since the square of every real number is positive we have our result.
Spencer
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@lhf, your version is much better. I didn't think to check the case possibility that $exp(x/2)=0$. This question should probably be closed as duplicate. – Spencer Apr 09 '14 at 14:45
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@Osama, your edit was inappropriate. Please refrain from trying to edit my answer again. If you have a recommendation to make then you may comment on my answer. – Spencer Apr 08 '17 at 20:32
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We have that $e>0$. Then for every $n\in \Bbb N$ we have that $e^n>0$.Because $\sqrt[m] .$ is increasing then $\sqrt [m] {e^n}>0$. Thus $$e^q>0$$ for every $q\in \Bbb Q$. Now $\Bbb Q$ is dense in $\Bbb R$ thus $e^x>0$ for every $x\in \Bbb R$.
Haha
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