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In 1914 Albert Bennett suggested the following operation:

$$a * b=a^0_2b=\exp(\ln a \ln b)$$

Now, given this function, addition and multiplication, and their properties, can one express exponentiation and power function?

The operation $a^0_2b$ has the following properties:

$$a*b =b*a$$ $$(ab) * c = (a * c)(b * c)$$ $$a * e = e * a = a$$

I also wonder whether a function $a^0_{-1}b=\log(\exp(a)+\exp(b))$ can help in this endeavor.

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    According to his 1915's paper, it seems that $a \# b = \exp(\log a + \log b)$, which actually denotes the usual multiplication.2012-11-06
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    By noting that the mapping $\log : (\Bbb{R}^{+},\#,*) \to (\Bbb{R},+,\cdot)$ gives an isomorphism, it seems natural to define the exponentiation via $$\log(a^{*b}) = (\log a)^{\log b}.$$2012-11-06
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    @sos440 I just used my own notation. If I had access to his paper I would change it.2012-11-06
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    The two operations, as @Anixx has quoted them, do not satisfy distributivity. So @sos440’s definition seems far more plausible. p.s. I met old Bennett during my first years at Brown.2012-11-06
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    It is not a matter of notation. Given definition of $\#$ and $*$, we cannot form an analogy of addition and multiplication since the distribution law $a * (b \# c) = (a*b)\#(a*c)$ fails. This is what I pointed out. And maybe you can refer to [*Note on an operation of the third grade*](http://www.jstor.org/stable/2007124?origin=crossref).2012-11-06
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    @Lubin there is no supposed distributivity, # is zero-order operation, before addition, and * is a 3rd order operation, after multiplication, it is distributive against multiplication.2012-11-06

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