Exponential function

glebovg · **Joined:** Mon, 28 Dec 2009 00:16:28 UTC **Posts:** 228

Let

be a positive irrational number and

be a positive rational number. Does ${e^s} = r$ hold only if $s = \ln (r)$ ? Are there any other values of

that make ${e^s}$ rational?

Shadow · **Posted:** Sun, 27 May 2012 04:19:17 UTC

glebovg wrote:

Let

be a positive irrational number and

be a positive rational number. Does ${e^s} = r$ hold only if $s = \ln (r)$ ? Are there any other values of

that make ${e^s}$ rational?

That would be the definition of an inverse function.

glebovg · **Joined:** Mon, 28 Dec 2009 00:16:28 UTC **Posts:** 228

Are there any cases when ${e^s}$ is rational? The Lindemann-Weierstrass theorem states that if $\alpha$ is algebraic, then ${e^\alpha }$ is transcendental, but when is ${e^s}$ rational?

outermeasure · **Posted:** Sun, 27 May 2012 04:26:54 UTC

glebovg wrote:

Are there any cases when ${e^s}$ is rational? The Lindemann-Weierstrass theorem states that if $\alpha$ is algebraic, then ${e^\alpha }$ is transcendental, but when is ${e^s}$ rational?

$\alpha=0$ ?

glebovg · **Joined:** Mon, 28 Dec 2009 00:16:28 UTC **Posts:** 228

But 0 is rational. Look at my original post. I defined

to be a positive irrational number.

Shadow · **Posted:** Sun, 27 May 2012 04:31:38 UTC

glebovg wrote:

But 0 is rational. Look at my original post. I defined

to be a positive irrational number.

No he meant that you are wrong, 0 is algebraic and e^0

is NOT transcendental, if you want to keep your original context you should say so.

In any case again, all you need is the definition of an inverse function.

glebovg · **Joined:** Mon, 28 Dec 2009 00:16:28 UTC **Posts:** 228

Hence the use of different letters. $\alpha$ is supposed to be a nonzero algebraic number. Sorry.

glebovg · **Joined:** Mon, 28 Dec 2009 00:16:28 UTC **Posts:** 228

So, there are no other rational numbers of the form ${e^s}$ ? Is there a theorem?

Shadow · **Posted:** Sun, 27 May 2012 04:51:04 UTC

glebovg wrote:

So, there are no other rational numbers of the form ${e^s}$ ? Is there a theorem?

I would not call that result "theorem", it's just immediate from the definition of an inverse function, or of a 1-1 function, or of a lot of other things.

glebovg · **Joined:** Mon, 28 Dec 2009 00:16:28 UTC **Posts:** 228

So, is ${e^e}$ irrational?

Shadow · **Posted:** Sun, 27 May 2012 05:31:08 UTC

glebovg wrote:

So, is ${e^e}$ is irrational?

unless e is the log of a rational number.

glebovg · **Joined:** Mon, 28 Dec 2009 00:16:28 UTC **Posts:** 228

What do you mean? Does this also work for all ${n^{{{\log }_n}r}}$ ?

Shadow · **Posted:** Sun, 27 May 2012 06:03:45 UTC

glebovg wrote:

What do you mean? Does this also work for all ${n^{{{\log }_n}r}}$ ?

Assuming you're dealing with positive, real n not equal to 1.

I mean, this is really basic stuff from algebra II in high school, look at the definition of the log function. If you still cannot convince yourself use the derivative to show the exponential (respectively log) are 1-1 on their domains, there are tons of ways to see this.

glebovg · **Joined:** Mon, 28 Dec 2009 00:16:28 UTC **Posts:** 228

But why is it unknown whether ${2^e}$ is irrational?

Shadow · **Posted:** Sun, 27 May 2012 07:01:12 UTC

glebovg wrote:

But why is it unknown whether ${2^e}$ is irrational?

Because that's equivalent to knowing if $\log_2(e)$ is rational...a very difficult thing to decide in general. Understanding which numbers we can easily express are or are not algebraic is a VERY difficult prospect in general.

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