Prove that $\log X < X$ for all $X > 0$ I'm working through Data Structures and Algorithm Analysis in C++, 2nd Ed, and problem 1.7 asks us to prove that $\log X < X$ for all $X > 0$.
However, unless I'm missing something, this can't actually be proven. The spirit of the problem only holds true if you define several extra qualifiers, because it's relatively easy to provide counter examples.
First, it says that $\log_{a} X < X$ for all $X > 0$, in essence.  
But if $a = -1$, then $(-1)^{2} = 1$. Therefore $\log_{-1} 1 = 2$. Thus, we must assume 
$a$ is positive.
if $a$ is $< 1$, then $a^2 < 1$.  Therefore we must assume that $a \geq 1$.
Now, the book says that unless stated otherwise, it's generally speaking about base 2 for logarithms, which are vital in computer science.
However, even then - if $a$ is two and $X$ is $\frac{1}{16}$, then $\log_{a} X$ is $-4$. (Similarly for base 10, try taking the log of $\frac{1}{10}$ on your calculator: It's $-1$.) Thus we must assume that $X \geq 1$.
...Unless I'm horribly missing something here. The problem seems quite different if we have to prove it for $X \geq 1$.
But even then, I need some help solving the problem. I've tried manipulating the equation as many ways as I could think of but I'm not cracking it.

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.
