Every graph with $n$ vertices and $2n$ edges has a cycle of length at most $2\log_2 n$? 
Prove that every graph with $n>1$ vertices and at least $2n$ edges contains a cycle of length at most $2\log_2 n$.

Since $K_n$ has $\binom n2=n(n-1)/2$ edges, it is clear that $n\geqslant 5$ for any graph with at least $2n$ edges, so the "with $n>1$ vertices" condition seems redundant. More importantly, $K_5$ has a cycle of length $5>2\log_25$, contrary to the claim. Perhaps I am misinterpreting the question?
 A: Yes this one indeed was a tough question to assign. The below is a sketch of the proof, make sure you can fill in the details.
Let $T$ be a tree rooted at $v_0$. Let $V_1,V_2,\ldots, V_{\ell}, \ldots$ be the set of vertices of $G$ where $V_{\ell}$ is the set of vertices of $G$ of distance precisely $\ell$. Now let $N_1,N_2,\ldots,N_{\ell}, \ldots$ be the number of vertices at distance precisely $\ell$ from $v_0$, for each $\ell < \log n$. So $N_{\ell} = |V_{\ell}|$
Fact 1: We can assume that the induced subgraph of $G$ on $V_1,\ldots, V_{\ell}$ for each $\ell \le \log n$ is a tree, otherwise we'd have a cycle no more than $2 \log n$ and we would be done.
Fact 2:We can also assume that for each such $\ell$ that the number of edges in $G$ with exactly one endpoint in $V_1 \cup \ldots V_{\ell-1}$ is the cardinality of $N_{\ell}$, otherwise there would also be a cycle of length $2 \ell$ or less (check to see if you can see why). So applying Fact 1, it follows that the number of edges incident to a vertex in $\{v_0\} \cup V_1 \cup \ldots \cup V_{\ell-1}$ is $|V_1 \cup \ldots \cup V_{\ell-1} \cup V_{\ell}|$, lest there is a cycle of length $2\ell$ or less.  (Why?)
Fact 3: Now let $\ell$ be less than $\log n$. Suppose that the inequality $N_{\ell} \le \sum_{i=1}^{\ell-1} N_{i}$ holds. Then Fact 2 that $U=\{v_0\} \cup V_1 \cup V_2 \cup \ldots V_{\ell}$ is incident to fewer than $2|U|$ edges. So throw $U$ out from $G$ and $G \setminus U$ will have more than $2n'$ edges where $n'$ is the number of remaining vertices.
So, Facts 1 through 3 imply that if there is no cycle of length $2 \log n$, then $\sum_{i=1}^{\ell} N_i > 2\sum_{i=1}^{\ell-1} N_i$ for each $\ell \leq \log n$. As there are only $n$ vertices though, this is impossible.
