cancelations and logarithms When faced with the problem of multiplying fractions, for example
$$
\frac 5 2 \cdot \frac 8 3\cdot \frac{9}{35}
$$
we know that we can permute the numerators, or equivalently, permute the denominators, getting
$$
\frac{5}{35}\cdot\frac 8 2 \cdot \frac 9 3
$$
and then cancel:
$$
\frac 1 7 \cdot \frac 4 1 \cdot \frac 3 1.
$$
Similarly when multiplying logarithms
$$
(\log_2 5)(\log_3 8)(\log_5 81)
$$
we can permute the arguments, or equivalently, permute the bases:
$$
(\log_2 5)(\log_3 8)(\log_5 81) = (\log_2 8)(\log_3 81)(\log_5 5)=3\cdot4\cdot1= 12.
$$
So we could say that in $(\log_2 5)(\log_3 8)(\log_5 81)$, we "cancel" the $5$s, getting $(\log_2 8)(\log_3 81)$.  Or that in $(\log_2 5)(\log_3 8)(\log_5 81)$ we "cancel" the $2$ and the $8$, getting $3(\log_3 5)(\log_5 81)$, and then "cancel" the base $3$ and the $81$, getting $3\cdot4\log_5 5$ and then "cancel" the $5$s, getting $3\cdot4\cdot1$.  Or that in $(\log_2 5)(\log_3 8)(\log_5 81)$ we "cancel" the $3$ and the $81$, getting $4\cdot(\log_2 5)(\log_5 8)$, and then "cancel" the $5$s, getting $4\cdot1\cdot\log_2 8$, etc.
However . . . . . . in the case of fractions, we can multiply numerators and multiply denominators, and say that
$$
\frac 5 2 \cdot \frac 8 3\cdot \frac{9}{35} = \frac{5\cdot8\cdot9}{2\cdot3\cdot35},
$$
so that we can say that in our cancelations, we are dividing both the numerator and the denominator of one fraction by the same thing.  Is there some way to do something analogous with logarithms and get something like $\log_{2,3,5} 5,8,81$, where the commas represent whatever operation is appropriate, which conceivably would be different in the base from what it is in the argument?
 A: About the question in your last two lines: not exactly, but pretty close if we first pass to one single common base. With your example:
$$\log_25\log_38\log_5 81=\frac{\log 5}{\log 2}\frac{\log 8}{\log 3}\frac{\log 81}{\log5}=\frac{\log 5}{\log 5}\frac{\log 8}{\log2}\frac{\log 81}{\log 3}=1\cdot3\cdot4=12$$
Here, "log" can be the natural one, the vulgar one or logarithm to any base.
A: The commas in $2,3,5$ can mean $\exp((\log 2)(\log 3)(\log 5))$, where the logarithms and the exponential function are both to the same base, and we don't care what base it is, and the commas in $5,8,81$ would mean $\exp((\log 5)(\log 8)(\log 81))$, where that same base is still used throughout, and then $\log_{2,3,5} 5,8,81$ really is the same thing as $(\log_2 5)(\log_3 8)(\log_5 81)$.  Dunno why I didn't think of this.
Later clarification in response to a comment below:
Say we let $x\circ y\circ z\circ\cdots = \exp_b((\log_b x)(\log_b y)(\log_b z)\cdots)$.  Then $$(\log_p q)(\log_r s)(\log_t u)\cdots =\log_{{}\,p\,\circ\,r\,\circ\,t\,\circ\,\cdots} (q\circ s\circ u\circ\cdots).$$
