Question: Just to establish notation with respect to the RSA protocol, let $n = pq$ be the product of two large primes and let $e$ and $d$ be the public and private exponents, respectively ($e$ is the inverse of $d \bmod \varphi(n)$). Given a plaintext message $m$, we obtain the ciphertext $c = m^e \bmod n$; we subsequently decrypt the ciphertext by calculating $c^d \bmod n$.

Suppose I'm trying to implement RSA on a device with low computational power, and these exponentiations take too long. I decide to make my implementation run faster by choosing small values for $e$ and $d$ (e.g. in the tens or hundreds). 

**Are there efficient attacks against such an implementation?**

Answer: 
First I must state that a *secure* RSA encryption *must* use an appropriate padding, which includes some randomness. See [PKCS#1](http://www.rsa.com/rsalabs/node.asp?id=2125) for details.

That being said, $d$ is the "private exponent" and knowledge of $d$ and $n$ is sufficient to decrypt messages. $n$ is public (by construction) so $d$ must be kept private at all costs. If it is very small then an attacker can simply try values for $d$ exhaustively.
On a more general basis, if the size of $d$ is lower than 0.29 times the size of $n$ (in bits) then there exists an efficient key recovery attack. The accepted wisdom is that trying to get a $d$ much smaller than $n$ is a bad idea for security.

On the other hand, there is no problem in having a small $e$, down to $e = 3$. Actually, with RSA as you describe, there *is* a problem with a very small $e$: if you use $e = 3$ and encrypt the very same message $m$ with three distinct public keys, then an attacker can recover $m$. But that's not really due to using a small $e$; rather, it is due to not applying a proper padding.
$e$ and $d$ *cannot* both be small, because $ed$ must equal at least $(p-1)(q-1)+1$, which is very large if the scheme is to provide any security against factoring algorithms.