# Real numbers as powers of real exponents

## Main Question or Discussion Point

Just a quick question for you guys, I've been unable to find the answer to this. Can all real numbers be written as n^p, where p is a real number?

What is n??

Anyway, negative numbers can't be written in that form.

Ok, let me restate. Can all positive real numbers x be written as n^p, where n and p are real numbers?

Or

if n < n * sqrt(n^2-1) < n^2, does that mean there exists p such that n * sqrt (n^2 -1) = n^p ?

Yeah. Take n=x and p=1. This even holds for negative numbers, so I guess my first reply wasn't quite correct.

Oh, mine wasn't complete, p needs to be different from 1.

The better question;

if n < n * sqrt(n^2-1) < n^2, does that mean there exists p such that n * sqrt (n^2 -1) = n^p ?

Take $n=\sqrt{x}$ and p=3.

Oh dear. Let me try this:

let S be a series from 2 to infinity of 1 / (n*sqrt(n^2 -1)).

Can you write S as 1/n^p?

if n < n * sqrt(n^2-1) < n^2, does that mean there exists p such that n * sqrt (n^2 -1) = n^p ?
This would force

$$p=\log_n{n\sqrt{n^2-1}}$$

This is ok if the logarithm exists. For that, we need n>0, n≠1 and $n\sqrt{n^2-1}>0$.

Also, if you're looking for an expression for p when the number n is fixed, then p is the logarithm (base n) of x. That is, given$$x=n^p$$you take logs on both sides,$$\ln x = \ln (n^p) = p \ln n$$and thus$$p = \frac {\ln x}{\ln n}$$so any positive x can be expressed this way (p will be negative for x<1, zero for x=1 and positive for x>1).

P.S.: Oops, two posts got in the middle while I was writing this. I was intending to continue after post #6.

Perfect, thank you!

Oh dear. Let me try this:

let S be a series from 2 to infinity of 1 / (n*sqrt(n^2 -1)).

Can you write S as 1/n^p?
This question makes no sense. In the series S, your variable n is the dummy variable and thus ranges over all positive integers. In your last sentence, n has become a fixed number.

My bad again, what I meant to say is can you write it as a series of 1/n^p.

It seems to me like you just want to check convergence of the series. Why can't you just do a comparison test with

$$\sum C\frac{1}{n^2}$$

where C is a constant.

Indeed I do, and yes I could do that. However, I'm interested in knowing if my approach is acceptable.

The problem with your approach is that (generally) your p will be dependent on n.

So if you follow Dodo's and my hints, then you won't end up with

$$\sum \frac{1}{n^p}$$

but rather with something in the exponent which is also dependent of n. This makes the situation harder.

Unfortunately, I used my approach on an exam, that's why I'm so keen on knowing if it's correct. Would have used the comparison, but alas, I didn't occur to me at the time.

Office_Shredder
Staff Emeritus