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.

I believe I managed to answer this question:
Proof is by contradiction:
Let us assume that Log X >= X for all X > 0.
Then we define Log X to be equal to Y.
X must be <= Y, by definition.
By the definition of logarithms, 2^Y = X.
Using simple substitution, we can say that 2^Y <= Y, since X is <= Y.
This inequality only holds true for Y < 0.  Since X <= Y, then the inequality also holds true only for X <= 0, via syllogism.  Thus Log X >= X if and only if X <= 0, thus Log X < X for all X > 0.
