$2^5 \cdot a^b=2,5ab$ I came across this problem in an elementary number theory book, and I think I solved it.  Well, the question is posed as 
$2^5 \cdot 9^2 = 2,592$.  Are there any other pairs $a,b \in \mathbb{Z}$ such that $2^5 \cdot a^b = 2,5ab$ is satisfied ($a,b$ are the tens and one representative, not a product)?  
So at first I just set up a $9\times9$ grid and sieved the answer (which is "No") mainly by eliminating such obvious answers as $a\ne1,2$, and $a^b\le82$.  Then I quickly realized that $2^5|25ab$, and there are only a select few integers where $32\cdot x=25ab$, for some integers $a,b,x$.  For example $32 \cdot 81 = 2,592 = 2^5 \cdot 9^2$, but this does not work for $32 \cdot 80=2,560 \neq 2^5 \cdot 6^0$ and $32 \cdot 79 = 2,528 \neq 2^5 \cdot 2^8$
So i think that, no, this is not possible except where $a=9, b=2$.
Now that I think i solved it, is there an algorithm for doing this quickly or does it require this sort of analysis for all $a^b \cdot c^d=a,bcd$?

There is a 'simple' way of doing this. Notice that $2^5\cdot a^b=25ab$ are both integers. Therefore, since $a^b$ is an integer, so too must be 
$$
a^b=\frac{25ab}{2^5}=\frac{25ab}{32}
$$
Now this can be done by hand very simply. We know that $2500<a^b<2600$. How many multiples of $32$ fit this inequality? This is simple to check by hand, the only possibilities are $32\cdot 79=2528$, $32\cdot 80=2560$, and $32\cdot 81=2592$. Now just factor the numbers:
$$
2528=2^5\cdot79^1\;\;,\;\;2560=2^9\cdot5^1\;\;,\;\;2592=2^5\cdot 3^4
$$
factoring out a $2^5$ yields the choices $79^1$, $5^1$, and $3^4$ for the value of $a^b$. Since $0\leq a \leq 9$, $79^1$ is eliminated. It is easy to check that $5^1$ cannot work for then $a=5$ and $b=1$ and the $2^5\cdot 5^1=160 \neq 2551$.
This leaves $a^b=3^4$. We can also write this $a^b=(3^2)^2=9^2$ or $a^b=81^1$. Since $a,b$ are integers with $0\leq a \leq 9$, either $a=3$ and $b=4$ or $a=9$ and $b=2$. We know the last to be true and the case of $a=3$ and $b=4$ is a simple matter to show doesn't work. This method does not require a computer or calculator and took only about $5$ minutes using simple pen and paper.
