I know that multiplying exponents of the same base will give you that base to the power of the sum of the exponents ($a^b \times a^c = a^{b+c}$), but is there anything that can be done with exponents that will give you some base to the power of the product of those exponents: $f(a^b, a^c) = a^{bc}$?

  • $\begingroup$ f(x,y)=xy, g(a,b)=a^b -> g(a,f(x,y)) $\endgroup$ – JMP Apr 7 '15 at 14:23
  • $\begingroup$ Is there context? That is, did this come up when you were trying to solve some other problem, or was it just something you were curious about? (Not that it matters much; I'm just curious.) $\endgroup$ – Akiva Weinberger Apr 7 '15 at 14:34
  • $\begingroup$ I was looking for an alternative, intuitive understanding of how combining e, pi, and i gets you -1, by trying to factor them into simpler expressions that can somehow be combined in some other way. I am familiar with other geometric explanations used for visualizing complex analysis. $\endgroup$ – Mark Cidade Apr 7 '15 at 14:37
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    $\begingroup$ Answerers are interpreting the question in different ways. Is $a$ fixed? If not, there is no hope. $\endgroup$ – Jonas Meyer Apr 7 '15 at 14:56
  • $\begingroup$ What do you mean by fixed? $\endgroup$ – Mark Cidade Apr 7 '15 at 15:04

The function as initially described fails to exist by the definition of a function: for any given input (or input "vector") $x$, there must exist a single output $f(x)$. Here, the quickest counterexample for any such function $f(a^b,a^c)=a^{bc}$ is to note that $2^4=4^2$ and then any combination of these produces multiple outputs for a single specified input:

$$f(2^4,2^8)=f(4^2,4^3)=f(16,16^\frac 32)=2^{32}\neq 4^6\neq 16^\frac 32$$

or more generally,

$$f(a^{bk},a^{ck})=f((a^k)^b,(a^k)^c)=a^{bck^2}\neq (a^k)^{bc}$$

Essentially, the problem is that with two inputs, our function does not have enough information to give us the result that we want. We must give the function an additional input, whether by specifying that input as a "parameter" that reduces $f$ to one member of a "family" of functions; or as a "variable" that is part of the input vector into $f$. Personally, since the $a$ referenced in the question looks a lot like it is intended to be a relatively "fixed" parameter, I would choose to solve the problem this way:

$$f_a(q,r) = a^{\log_a q\cdot\log_a r}$$

where you are declaring the function $f$ to be bound by the base $a$ that you intend to work in. Now, we would give our function some inputs like $f_a(a^b,a^c)$ and expect to get a value $a^{bc}$.

With the additional comments discussing the possibility of using this function to understand the quantity $e^{i\pi}$ by applying our function to $f_e(e^i,e^\pi)$, we would naively apply our function as

$$f_e(e^i,e^\pi)=e^{\log e^i\cdot \log e^\pi} = e^{i\pi}$$

and arrive at the correct answer, but this process fails to provide further meaning or understanding, and also ignores the possibility that our function may take on other values when supplied with complex inputs outside of what we intend.

  • $\begingroup$ This doesn't seem to give correct results as $ f_2(4,4)$ should equal 65536 not 16... $\endgroup$ – Zach466920 Apr 7 '15 at 14:42
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    $\begingroup$ @Zach466920: it can be defined either way, depending on whether $q,r$ are the original $a^b,a^c$ or $b,c$ $\endgroup$ – Ross Millikan Apr 7 '15 at 14:58
  • $\begingroup$ @RossMillikan a is the base, which is 2. q and r equal 4. The result should be $2^{4*4}$ as asked for by the Op. Is there some ambiguity I'm missing? $\endgroup$ – Zach466920 Apr 7 '15 at 15:02
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    $\begingroup$ This answer is an $f$ such that $f(a^b,a^c) = a^{b * c}$, not such that $f(b,c) = a^{b*c}$. $\endgroup$ – JHance Apr 7 '15 at 15:14
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    $\begingroup$ @MJD: you're welcome, it's the one I wish I'd started with too, but "tyranny of the urgent" and all that... $\endgroup$ – abiessu Apr 8 '15 at 15:24

There is not. Consider as one example $$f(2^4, 2^4) = 2^{4\cdot 4} = 2^{16} =65536.$$ And you also want $$f(4^2, 4^2) = 4^{2\cdot 2} = 4^4 = 256.$$

But the left sides are equal (because $2^4=4^2$) while the right sides are not. So the arguments of the function do not have enough information to produce the results you want.

  • $\begingroup$ As already mentioned this isn't a requirement. You're noting that equality doesn't hold, but you forget that the operator operates on parts of the operand, namely the power its raised to. It should be expected for this to happen... $\endgroup$ – Zach466920 Apr 7 '15 at 14:53
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    $\begingroup$ Operators don't work that way. $\endgroup$ – MJD Apr 7 '15 at 14:54
  • $\begingroup$ An operator that only operates on the power a number is raised to. Seems like the floor/ceiling/int operators that only operates on the whole part of a number. $\endgroup$ – Zach466920 Apr 7 '15 at 14:56
  • $\begingroup$ Just thought of this. Order of operations also justifies this. You"re saying the power operations happens before the new one. I say the new operation happens first. $\endgroup$ – Zach466920 Apr 7 '15 at 15:15

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