Integer functions For $x>0$ consider the following three functions: 
$$\begin{align}
f(x)&=x+1;\\g(x)&=2x;\\t(x)&=3x
\end{align}$$
Let $A(x)$ be the minimum number of operations using only functions $f(x)$ and $g(x)$ needed to get $X$ from $0$.
Let $B(x)$ be the minimum number of operations using only functions $f(x) $ and $t(x)$ needed to get $X$ from $0$.

Endless or no set of numbers for which $A(x)<B(x)$.
 A: To expand on Antony's suggestion, for a number $x$, $A(x)$ is simply the sum of the digits of the binary representation of the number, plus the number of digits $\lfloor log_2(x) \rfloor$. Similarly, $B(x)$ is the sum of the digits of the ternary representation of the number, plus the number of digits $\lfloor log_3(x) \rfloor$.
The expected value of $A(x)$ and $B(x)$ for a "random" integer would be:
$E(A(x)) = \frac{3\lfloor log_2(x) \rfloor + 1}2$ and 
$E(B(x)) = 2 \lfloor log_3(x) \rfloor + 1$,
which means that with $x$ taken to infinity $\frac{E(A(x))}{E(B(x))} = \frac{3 log_2(3)}4 \approx 1.1887...$
Therefore it is to be expected that $A(x)$ is usually larger than $B(x)$. However, statistically any number which has a binary representation with less $1$'s than $2 \lfloor log_3(x) \rfloor + 1 - \lfloor log_2(x) \rfloor$, which comes out roughly $(\frac2{log_2(3)}-1)log_2(x) + 1 \approx 0.26186log_2(x) + 1$. Meaning that if a number has in its binary expansion roughly a quarter or less of the digits (excluding the highest digit, which we eliminate with the $+1$ as it must be $1$) as $1$s, then chances are $A(x) < B(x)$. This means that as $x \to \infty$, the number of integers with $n$ digits that we can expect $A(x) < B(x)$ is approximately $\binom{n}{n/4}$, which isn't very much compared with $2^n$, but still tends to infinity.
That said, in order to answer the question, we need to try and find an infinite set of $x$ which we can be sure $A(x)<B(x)$. The powers of 2 suggested by mjqxxxx appear to be good candidates because they minimize $A(x)$, however I'm not sure how you can prove $B(x)$ is bigger even for this sequence (some numbers may by very small chance have a very low $B(x)$ value as well).
