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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?

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I'm not precisely sure if this addresses your question, but isn't the logarithm cancelling method precisely the same as the fraction cancelling once you read $ \log_a b = \dfrac{\ln b}{\ln a}$ ? –  Ragib Zaman Oct 3 '12 at 3:15
    
@RagibZaman : The identity at the end of your comment is of course the basis of this whole thing, but I don't understand how it means that it's "precisely the same thing". –  Michael Hardy Oct 3 '12 at 3:18
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2 Answers 2

up vote 3 down vote accepted

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.

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I think you've very nearly got it. The commas in $2,3,5$ can mean $\exp((\log 2)(\log 3)(\log 5))$, where the logarithm and 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. –  Michael Hardy Oct 3 '12 at 3:28
    
Indeed so, @MichaelHardy. –  DonAntonio Oct 3 '12 at 3:40
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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 $$(\log_p q)(\log_r s)(\log_t u)\cdots =\log_{{}\,p\,\circ\,r\,\circ\,t\,\circ\,\cdots} (q\circ s\circ u\circ\cdots).$$

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It's not clear to me what you mean by the above. Could you please elaborate. –  Bill Dubuque Oct 4 '12 at 20:59
    
Say we let $x\circ y\circ z\circ\cdots = \exp_b((\log_b x)(\log_b y)(\log_b z)\cdots)$. Then $(\log_p q)(\log_r s)(\log_t u)\cdots$ $=\log_{{}\,p\,\circ\,r\,\circ\,t\,\circ\,\cdots} (q\circ s\circ u\circ\cdots)$. That's what I meant. –  Michael Hardy Oct 4 '12 at 21:54
    
So, was your goal simply to find an operation $\,\circ\,$ such that $$\rm log_{\ a\circ b\circ c}(x\circ y\circ z)\ =\ \frac{log(x\circ y\circ z)}{log(\ a\circ b\circ c)}\ =\ \frac{log(x)\,log(y)\,log(z)}{log(a)\,log(b)\,log(c)}$$ If so, it would help to edit your question to make that more clear. –  Bill Dubuque Oct 4 '12 at 22:10
    
Certainly that was not my goal, since (as I said in the comments below the answer I "accepted") I hadn't even noticed the obvious fact pointed out by "DonAntonio" in that answer. But I wrote "where the commas represent whatever operation is appropriate", so identifying that operation if it existed was at least a part of the question. –  Michael Hardy Oct 4 '12 at 22:14
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Ok, so what was your original problem? That's the best sense I can make of it so far. –  Bill Dubuque Oct 4 '12 at 22:16
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