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When learning the laws of exponents and logarithms, one finds that there is a correspondence. Each law of exponents has a corresponding equivalent expression in terms of logarithms. For example, the exponential of a sum law, $a^{x+y}=a^xa^y$ becomes $\log(wz)=\log w + \log z$ after you take a logarithm and relabel. Or in words, since exponentials turn sums into products, their inverse function must turn products into sums. The Repeated exponentiation law $(a^x)^y=a^{xy}$ becomes $\log z^y = y\log z$.

If you make a list of all the rules of exponents, you'll find that all of them (with one exception) has a useful form expressed in terms of logarithms. Personally I find understanding this relationship between the two sets of identities to be the best way to learn them, remember them, and teach them.

However there is one law, which may be called the distributive law for exponentiation, or the power of a product law, namely $(ab)^x=a^xb^x,$ for which there seems to be no useful logarithmic version. My question is, why not? I can see that because it involves exponentiation operations with three different bases, if there were a logarithmic identity it would awkwardly involve different logarithm bases. But is there more we can say?

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    $\begingroup$ The direct equivalent is $x \log(ab) = x \log(a) + x\log(b)$, but this is just a restatement of $\log(ab) = \log(a) + \log(b)$. $\endgroup$
    – Dan
    Commented Jun 14 at 18:31
  • $\begingroup$ Writing $b=a^q$ (where $q=\frac{\log b}{\log a}$), the distributive law looks like $(a\cdot a^q)^x = a^x (a^q)^x$; one can verify that this is a consequence of the first two laws. $\endgroup$ Commented Jun 14 at 19:33

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