For all reals $x$, prove $2^x > x$ How can I prove that for all reals $x$, $2^x > x$? I can prove this for integers with induction, but I can't figure out how to prove it for reals. Perhaps you could say that since $2^x$ is strictly increasing, $2^x \ge 2^{floor(x)} > floor(x)$, but is this really the best approach?
 A: Can you show that $x\log2>\log x$, $\, \forall\, x>0$?
A: Since you have proven it for all integers, the key step has been done. Now note that we are only to show the inequality for $x \in \mathbb R^+$ because the inequality becomes trivial for $x \in \mathbb R^-$ and when $x=0$, the inequality is obviously true. So now we note that $2^n>n$ $\forall n$ $\geq 0$. In other words we have $2^n \geq n+1$ $\forall n$ $\geq 0$. Now choose any $\epsilon$ such that $n<\epsilon<n+1$. We then get $2^n$$^+$$^1$ $>$ $2^\epsilon$ $>$ $2^n$ $\geq$ $n+1$ $>$ $\epsilon$ $>$ $n$. Hence proved for all positive reals.
Note that I have claimed that the result is proved for all positive reals but has not shown that why it must be. Prove this and the result will be completely proved. 
A: Hint
The derivative of the function $f(x) = 2^x-x$ is $$f'(x) = \ln(2) \cdot 2^x - 1$$ which has only one zero (let's say at $x_0$). Check that this is a minimum point and make sure that $f(x_0)>0$, and you are done, since $f(x)$ is continous.
A: Hint
Take the $^2\log$ on both sides
