1

How does one prove $\log{x} < x^n$ for all $n > 0$.

Formally, I would like to show that for all $n > 0$, there exists a $Y$ after which for all $x>Y$, $\log{x} < x^n$.

From there, is it then possible to show $\log{(x)} \cdot \log{(\log{(x)})} < x^n$?

(Unrelated, and just interesting:) In general, is the product $\log(x) \cdot \log(\log(x)) \cdot \log(\log(\log(x))) \cdots < x^n$?

Arturo Magidin
  • 398,050
  • I'm sorry I think your edit crossed over with mine so the colons seem to be back. But I don't want to edit it anymore since too many edits will turn your question into community wiki, I believe, and it seems that there are 6 edits already. – Adrián Barquero Mar 29 '11 at 05:58

5 Answers5

3

Here's yet another way, if you may assume that the inequality $0 < \ln t < t-1$ for $t>1$ is known. (If you define $\ln t$ as the area under the graph $y=1/x$ from $x=1$ to $x=t$, it's clear that this area is smaller than the area of a rectangle of height one and width $t-1$.)

To simplify a little, one can use the weaker inquality $0 < \ln t < t$, or $$ 0 < \frac{\ln t}{t^2} < \frac{1}{t} $$ for $t>1$. Now let $t = x^{n/2}$ (where $n>0$ is not necessarily an integer): $$ 0 < \frac{(n/2) \ln x}{x^n} < \frac{1}{x^{n/2}} $$ for $x>1$. The right-hand side tends to zero as $x \to \infty$, so by the squeeze theorem you get $$\lim_{x \to \infty} \frac{\ln x}{x^n} = 0,$$ which implies what you wanted to show.

Hans Lundmark
  • 53,395
2

Hint: Try looking at the derivatives. If $\frac{d}{dx}(\text{log}(x)) < \frac{d}{dx}(x^n)$ for all $n > 0$, what does this tell you about the two functions?

Alex Becker
  • 60,569
  • 1
    $\frac{d}{dx} (1-\frac{1}{x}) \gt \frac{d}{dx} (1-\frac{1}{x^2})$ for $x \gt 2$ and yet $1-\frac {1}{x} \lt 1-\frac{1}{x^2}$ on that range. – Douglas Zare Mar 29 '11 at 08:28
1

Once you show that for all $c>0$, $\log(x)<x^c$ for sufficiently large $x$, it follows that for all $b>0$ and $c>0$, $\log(x)^b<x^c$ for all sufficiently large $x$, by applying the first result with exponent $c/b$. Notice that $\log(\log(x))<\log(x)$, which implies that $\log(x)\cdot \log(\log(x)) < \log(x)^2$. Similarly, all of your finite products of logs are less than some power of $\log(x)$, and the result follows.

Jonas Meyer
  • 53,602
0

Let $g(x) = x^{n} - logx $ for $x > 0$.

The first derivative is $g^{(1)}(x) = n x^{n-1} - 1/x$

It follows that $g^{(1)}(x) > 0 => x > 1$ , $ g^{1}(x) = 0 => x = 1 $ , $ g^{1}(x) < 0 => x < 1 $

Therefore, x=1 is a local minimum. However , $g(1) =1$ , therefore $g(x) > 0 \forall x>0$

chazisop
  • 340
0

Note, that this can only be true if $n\ge1$ or the like, because for instance $2\gt 2^{0.0000001} $
Also it seems we have to assume $x\gt0$

substitute $y=\log(x)$ as in some previous examples here on stackexchange

$ \qquad y < \exp(y)^n =\exp(y*n) $
$ \qquad y < 1+ (y*n) + {(y*n)^2 \over 2! } + ... $ which is then obvious based on comparision of terms.

Gottfried Helms
  • 34,920