Proving that $f(x)=2^x$ is $O(x^2)$ Can someone help me with this problem? I don't really know what to do if the x is in exponential form. 

To see that this is way false, we just observe:
$$2^x\le Mx^2\iff x\log 2\le \log M+2\log x$$
$$\iff x\le \log_2(M)+2\log_2(x)$$
But it is clear from L'Hôpital's rule (among other things) that this is false, since
$$\lim_{x\to\infty} {M\over x\log 2}+{2\log x\over x\log 2}=\lim_{x\to\infty} {2/x\over \log 2}=0$$

(in fact this would imply)
$$x=O(\log x)$$
