Why didn't Frege succeed in his attempts to reduce mathematics to logic? My background: Sophomore-level understanding of mathematics and philosophical logic.
All the explanations I have found online so far are either far too technical or too simplistic.
Thanks in advance for your help! :-)
 A: Frege's 'logic', which he used informally in Foundations of Arithmetic and more formally in Basic Laws of Arithmetic was essentially what would be recognised nowadays as second-order logic (with full comprehension) together with a single axiom, which he called Basic Law V. This says:

(BLV) The extension of the concept $F$ = the extension of the concept $G$ if and only if everything that is $F$ is a $G$ and vice-versa.

or symbolically:
$$ \forall F\forall G ( \{x:Fx\} = \{x:Gx\} \leftrightarrow \forall x(Fx\leftrightarrow Gx) ) $$
It's sufficient for understanding that a concept for Frege is essentially something like a property, and an extension of a concept is essentially the set of all things satisfying that property. As a consequence of the second-order comprehension scheme*, for every open formula, there is a corresponding concept. This together with BLV entails a principle of naive set comprehension:
$$\exists x\forall y(y\in x \leftrightarrow \phi(y)) $$
for any formula $\phi$. (Given Frege's notation, we need to define the membership relation, rather than take it as primitive, as is the standard approach now.)
From his logic (which includes BLV), Frege succeeded in developing a significant amount of arithmetic.
There are, however, two problems with the 'logic' in which he does this (the second of which is significantly more problematic than first).
First, BLV isn't really a principle of logic, it's an axiom, similar to, say, the axiom of extensionality in modern set theory.
Second, naive comprehension is inconsistent, since it allows one to derive Russell's Paradox (which Russell communicated in a letter to Frege). If we consider the formula
$$ x\notin x $$
then, by naive comprehension, there is a set of all and only those things which satisfy this formula:
$$ \exists x\forall y(y\in x \leftrightarrow y \notin y) $$
Let's call this set $r$. But then, $r\in r$ iff $r\notin r$, which is a contradiction!
Frege tried to solve this problem by modifying BLV. Unfortunately, his suggested modification is still inconsistent (see, for example, here). 
.* Actually, Frege did not have a comprehension scheme, but a principle of substitution which has the same effect. There's a good explanation of the relation here.
