Wolfram Mathematica simplifies $(a^b)^c$ to $a^{bc}$ only for positive real $a, b$ and $c$. See W|A output.

I've previously been struggling to understand why does $\dfrac{\log(a^b)}{\log(a)}=b$ and $\log(a^b)=b\log(a)$ not always hold (while I was always thinking of logarithms of positive reals as $\log_a(a^b)=\dfrac{\log(a^b)}{\log(a)}=\dfrac{b\log(a)}{\log(a)}=b$ before I've started to self-study complexes), but then learned about branch cutting of multifunctions, so that natural logarithm can be defined to be a function by first finding a set of $w$ for each $z$, so that $z=e^w$ (inverse natural exponential), and then somehow (no matter how) selecting a unique solution $w$ from that set, so that $\forall{z}\exists!{w}$, yielding a function $z\mapsto{w}$. This results in some sort of discontinuity where solutions are being dropped: branch cut

This is done because functions are generally more easy to deal with than multifunctions. Wolfram's convention is to define $\log(z)$ as the inverse of $e^z$ such that $\log(1)=0$ and such that the branch cut discontinuity is at ${(-\infty;0]}$. I know this is somewhat unpopular to think of natural logarithm as a single-valued function, but I am going to follow this convention (I personally think it is well-justified, at least that's what happens with $\operatorname{arcsin}$, $\operatorname{sqrt}$, etc, so in fact I am comfortable with it).

Then it is clear for me why is $\dfrac{\log\left((-2)^{-3}\right)}{\log(-2)}\approx{0.814319+0.841574i}\neq{-3}$, while $-3$ perfectly satisfies the equation $(-2)^x=(-2)^{-3}$.

Given that $\log(a^b)=b\log(a)$ holds for positive real $a, b$ only, I thought how would I then solve equations of form $a^x=b$, where $a,b$ are complexes and not necessarily positive reals, since my first step was always to rewrite everything to base $e$: $$e^{\log(a^x)}=e^{\log(b)}$$ and then carry the exponent out of the logarithm: $$e^{x\log(a)}=e^{\log(b)}$$ (adding $2\pi i k,\,k\in\mathbb{Z}$ to any exponent and then eliminating the exponentials yields the result). So I've asked on ##math, where I've been pointed out that another way to rewrite $a^b$ to base $e$ is using the definition of the logarithm: $a^b=(a)^b=(e^{\log(a)})^b=e^{b\log(a)}$. I was happy with that (and even solved a $(-2)^x=-3$ for $x$ just for fun using this approach, see here) but some time later I have realized that this neither does not actually work for arbitrary complex $a$ and $b$! $$1=1^{1/2}=((-1)^2)^{1/2}=(-1)^{2\cdot\frac{1}{2}}=(-1)^1=-1$$ The rule $(a^b)^c$, as I was told by W|A, requires $a, b, c$ to be positive reals. When I pointed that out on ##math, I was told that $a^b=e^{b\log(a)}$ is by definition of complex exponentiation. I've checked, and Wolfram Mathematica agreed with this identity. Too good! But then I realized from these something must not be true:

  • $(a^b)^c=a^{bc}$ holds only for positive real $a,b,c$
  • $\forall{a,b\in\mathbb{C}}:a^b=e^{b\log(a)}$
  • $\forall{a\in\mathbb{C}}:a=e^{log(a)}$

The latter is the definition of the logarithm, so should be true. The second also must be true, otherwise I do not know how to solve equations. Hence: $$a^c=e^{c\log(a)}$$ is by the definition of complex exponentiation, just as I was told. Then, rewriting $a$ in the LHS using the definition of logarithm: $$a=e^{log(a)}$$ $$\left(e^{log(a)}\right)^c=e^{c\log(a)}$$ Now, since that for every complex $b$ there exists a $z$ such that $b=\log(z)$, we can rewrite $\log(a)=b$: $$(e^b)^c=e^{bc}$$ for all complex $b,c$! So, what is that?

I've made a mistake? Or is base $a=e$ that special?

Or is in fact (I suspect) one just needs to require $a$ to be positive real, and $b,c$ are in fact irrelevant? $$\forall{a}\in\mathbb{R}\forall{b,c}\in\mathbb{C}:a>0\implies{(a^b)^c=a^{bc}}$$ Have I found a bug in Mathematica and W|A or made a huge stupid mistake leading myself to drastic misunderstanding?

P. S. This is my first post at MSE, I am not a math major, just a hobbyist, so sorry if I am struggling at basics here. Also sorry for my English: it is not my native language.

Edit: thank you @Andrew for your answer.


Very clear and straightforward, works flawlessly.

But it appears that though the implication is obviously true, there are more cases (read "values of $a,\,b,\,c$") from which $(a^b)^c=a^{bc}$ does follow, i.e. I found that it is true for $c\in\mathbb{Z}$ and arbitrary complex $a,\,b$, for example:

$$\big((-2)^{-3}\big)^{-2}=(-2)^{(-3)\times(-2)};$$ $$\left(\frac{1}{(-2)^3}\right)^{-2}=(-2)^{6};$$ $$\left(-\frac{1}{8}\right)^{-2}=64;$$ $$\left(-8\right)^2=64;$$ $$64=64.$$

For this case, $b\log(a)=-3\log(-2)=-3(\log(2)+i\pi)=-3\log(2)-3i\pi$, and hence $\Im(-3\log(2)-3i\pi)=-3\pi\notin{({-\pi;\pi}]}$, therefore @Andrew's implication does not cover all cases.

So, is there any more solutions of $(a^b)^c=a^{bc}$?

  • $\begingroup$ Just saw your update; probably the "@" trick only works in comments? In my answer, the stated claim is "If $b \log a$ has imaginary part between $(2k - 1)\pi$ and $(2k + 1)\pi$, then ... $(a^{b})^{c} = a^{bc}$ if and only if $\exp(2\pi cki) = 1$." The example in your edit ($c$ an integer) fits this claim. :) Other examples would include $c = p/q$ a rational number in lowest terms and $k$ an integer multiple of $q$. $\endgroup$ – Andrew D. Hwang Aug 20 '15 at 16:10
  • $\begingroup$ I am sorry @Andrew I do not understand what do you say. If $k$ should be fixed so that the range is equal to the range of the imaginary part of complex logarithm of selected branch, then, because I set the branch as $\Im\left(\log(z)\right)\in{({-\pi};{\pi}]}$, I fix $k=0$. Then $\exp(2\pi cki)=\exp(0)=1$ for any $c$, hence you literally claim that $(a^b)^c=a^{bc}$ is always true. $\endgroup$ – dbanet Aug 20 '15 at 16:28
  • 1
    $\begingroup$ Oh. I've got it. Forget my previous comment. In this case $\Im(b\log(a))=-3\pi$, so from here it follows that $k=-1$, then $\exp(2\pi(-2)(-1)i)=\exp(4\pi i)=1$. $\endgroup$ – dbanet Aug 20 '15 at 16:56
  • $\begingroup$ By the way @Andrew, it does not matter which side of the interval $(2k\mp1)\pi$ is inclusive, can be both, does it? $\endgroup$ – dbanet Aug 20 '15 at 17:19
  • $\begingroup$ "It doesn't matter" with some fine print: If we fix the imaginary part of $\log$ to lie in $(-\pi, \pi]$, then $k$ is defined by "the imaginary part of $b \log a$ is in $\bigl((2k - 1)\pi, (2k + 1)\pi\bigr]$", and the argument is as stated. If the imaginary part of log lies in $[-\pi, \pi)$, the interval defining $k$ gets a corresponding shift of endpoint. That is, shifting the value of $\log$ on the negative reals affects the definition of exponentiation, but "in a consistent way" with the stated conclusion. $\endgroup$ – Andrew D. Hwang Aug 20 '15 at 17:33

If we agree $\log z$ has imaginary part between $-\pi$ and $\pi$ and is defined only on the set $D = \mathbf{C} \setminus(-\infty, 0]$, then \begin{align*} \exp(\log z) &= z\quad\text{for all $z$ in $D$,} \\ \log(\exp z) &= z\quad\text{for all $z$ with imaginary part between $-\pi$ and $\pi$.} \end{align*} If the imaginary part of $z$ is between $(2k - 1)\pi$ and $(2k + 1)\pi$, then $$ \log(\exp z) = z - 2\pi ki \tag{1} $$ because $z - 2\pi ki$ has imaginary part between $-\pi$ and $\pi$.

Defining $a^{b} = \exp(b \log a)$, we have \begin{align*} (a^{b})^{c} &= \exp\bigl(c \log(a^{b})\bigr) = \exp\bigl(c \log [\exp (b \log a)]\bigr), \\ a^{bc} &= \exp(bc \log a). \end{align*}

If $b \log a$ has imaginary part between $(2k - 1)\pi$ and $(2k + 1)\pi$, then $\log\bigl(\exp(b \log a)\bigr) = b \log a - 2\pi ki$ by (1), so $$ \exp\bigl(c \log(a^{b})\bigr) = \exp\bigl(c(b \log a - 2\pi ki)\bigr) = \exp(bc \log a) \exp(-2\pi cki), $$ which is equal to $a^{bc}$ if and only if $\exp(2\pi cki) = 1$.

In particular, if $b \log a$ has imaginary part between $-\pi$ and $\pi$ (i.e., $k = 0$), or if $c$ is an integer, then $$ (a^{b})^{c} = \exp\bigl(c \log(a^{b})\bigr) = \exp\bigl(c \log(\exp b \log a)\bigr) = \exp(bc \log a) = a^{bc}. $$

  • $\begingroup$ Um... $\log$ is defined on ${(-\infty;0)}$, and is also defined at $0$ if we allow the point at infinity. For example, $\log(-2)=\log(2)+\pi i$, $log(0)=-\infty$. $\endgroup$ – dbanet Jul 2 '15 at 23:11
  • $\begingroup$ Hah, FullSimplify[Log[e^z],Assumptions→-π<Im[z]≤π] indeed yields z in Mathematica! $\endgroup$ – dbanet Jul 2 '15 at 23:27
  • $\begingroup$ Cool, I think I'm getting it. $\endgroup$ – dbanet Jul 2 '15 at 23:54
  • $\begingroup$ If $\log(-a) = \log a + i\pi$ for $a > 0$ real, the argument goes through with obvious modifications. :) (Trying to make sense of things for $a = 0$ isn't worth the hassle; there's no notion of signed infinities in $\mathbf{C}$, and $\exp$ has an "essential singularity" at $\infty$, so defining $\log 0 = \infty$ is asking for trouble.) $\endgroup$ – Andrew D. Hwang Jul 3 '15 at 0:38
  • $\begingroup$ Disagree. While considering plain $\mathbb{C}$, there indeed is no notion for any infinities, but compactifying the complex plane to a sphere using one additional point, the point at infinity, yields Riemann sphere, for which signed infinities, and even more, are possible: consider a vector connecting the origin and the point at infinity. Obviously, there are infinitely many such vectors, differing with theirs angles. Well, generally there are just two points for which the angle of the vector connecting the origin with the point is not unique and well defined: $\endgroup$ – dbanet Jul 3 '15 at 3:31

This is a late post. But just to make it clear for future readers, the error in your question is here:

... for every complex $b$ there exists a $z$ such that $b=\log(z)$, ...

This is false since you already chose $\log$ to be a function, and you cannot choose to have both $\log(1) = 0$ and $\log(1) = 2πi$, which implies that either $0$ or $2πi$ will not be a possible value of $\log$ since $\exp(0) = 1 = \exp(2πi)$. To put it algebraically, it cannot be that we have both $\log(z) = b$ and $\log(w) = b+2πi$ for some complex $b,z,w$, otherwise $z = \exp(b) = \exp(b+2πi) = w$ and hence $\log(z) = \log(w)$, contradicting $b \ne b+2πi$.


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