Prove that $\log X < X$ for all $X > 0$

Solution 1:

One way to approach this question is to consider the minimum of $x - \log_a x$ on the interval $(0,\infty)$. For this we can compute the derivative, which is $1 - 1/(\log_e a )\cdot x$. Thus the derivative is zero at a single point, namely $x = 1/\log_e a,$ and is negative to the left of that point and positive to the right. Thus $x - \log_a x$ decreases as $x$ approaches $1/\log_e a$ from the left, and then increases as we move away from this point to the right. Thus the minimum value is achieved at $x = 1/\log_e a$. (Here I'm assuming that $a > 1$, so that $\log_e a > 0$; the analysis of the problem is a little different if $a < 1$, since then for $x < a < 1$, we have $log_a x > 1 > x,$ and the statement is not true.)

Now this value is equal to $1/\log_e a + (\log_e \log_e a)/\log_e a,$ and you want this to be $> 0 $. This will be true provided $a > e^{1/e}$ (as noted in the comments).

Solution 2:

Suppose that $x>0$ and $\log x > x$. Because the exponential map is monotonic increasing, it follows that $x > e^x$. This contradicts the well-known series representation for $e^x$:

$$x\not>\sum_{k=0}^\infty \frac{x^k}{k!}=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots$$

We also have the following, stronger result: since $e^x>1+x$ for all $x>0$, we have $x > \log(1+x)$ for $x>0$.

Edit: It appears that the question concerns $\log_2$ and not $\log$, which requires a different proof. Thus, for which $b >1$ do we have $\log_b x <x$ for all $x>0$?

The proof above shows that the set of admissible $b$ is non-empty. As $b \to 1$ from above, the curves $\log_b x$ sweep down towards the line $f(x)=x$. It's not hard to see that these functions sweep past $f(x)$, hence there exists a maximal $b$ such that $\log_{b} x \not < x$ for all $x>0$. It follows that $\log_{b} x$ share a tangent line for some unique $b_0$.

Let $x_0$ denote such a point of tangency. Comparing slopes, we must have $1/(x_0\log b_0)=1$, i.e $x_0=1/\log b_0$. As the functions $\log_{b_0} x$ and $x$ must moreover agree at $x_0$, we compute $\log_{b_0}(1/\log b_0)=1/\log b_0$, hence $\log b_0=1/e$. This gives a value of $$b_0 =e^{1/e} \approx 1.44467,$$ such that $b>b_0$ implies that $\log_b x < x$ for all $x >0$. In particular, $b_0<2$, so that $\log_2 x < x$ for all $x>0$.

There may be some details swept under the rug, but the idea (and resulting constant) is certainly correct.