Is it possible to represent any positive integer with a sum of arbitrarily many distinct powers of $3$, $5$, and $7$? My question is whether it is possible to represent every positive integer as the sum of any number of unique terms $b^x$, where $b\in\{3,5,7\}$, and integer $x \geq 0$.
(Note this would be trivially easy if using the prime base $2$ were allowed, as using additive terms $2^i$ is essentially how binary numbers are written.)
For example, in my scenario, $24$ has two valid representations:
$$ 3^1+3^2+5^1+7^1 = 3^0+3^2+5^0+5^1+7^0+7^1 = 24 $$
My instinct is that this shouldn't hold, but I haven't been able to find a counterexample yet or a convincing argument against it. I have been able to find empirical evidence that it doesn't work for almost all larger triples, e.g. $\{3,7,13\}$.

Edit
While Brian provided a nice counterexample to my $\{3,5,7\}$ case, it's natural to ask whether the same result will hold for $\{3,5,7,11\}$, or in general, for any finite subset of odd primes. Also remaining is the question of how to effectively determine the minimum counterexample for a given set.

$3^{60} - 1$ isn't achievable.  
This can be seen as 
$$\lfloor\log_5(3^{60}-1)\rfloor = 40 \\ \lfloor\log_7(3^{60}-1)\rfloor = 33$$
but
$$\sum_{n=0}^{59} 3^n + \sum_{n=0}^{40} 5^n + \sum_{n=0}^{33} 7^n < 3^{60} - 1$$

Edit: Please also see the generalization given by user687721 in the comments.  If you can understand what is said there, you'll see what I did here is just a specific example of that procedure, and the resulting test given in the comment could be done by a grade-schooler with a pencil (whereas I needed a computer, or at least a decent calculator, for several steps)
