I'm so puzzled about this:

$$a^{b^c} = a^{(b^c)} \neq (a^b)^c = a^{(bc)}.$$

Why isn't $a^{b^c}$ equal to $a^{(bc)}$? Why is $a^{b^c}$ instead equal to $a^{(b^c)}$? And how is it possible that $(a^b)^c = a^{(bc)}$?

My mind is pretty much exploding from trying to understand this.

  • $\begingroup$ full blown real numbers $\endgroup$ – bodacydo Nov 18 '12 at 14:15

That $a^{b^c}$ stands for $a^{(b^c)}$ rather than for $(a^b)^c$ is merely a notational convention; we say that the exponentiation notation associates to the right (whereas arithmetic operations associate to the left, so that $a-b-c$ means $(a-b)-c$ rather than $a-(b-c)$).

The fact that $(a^b)^c=a^{b\times c}$ is easy to understand: $a^b$ is obtained by multiplying together a sequence of $b$ copies of $a$, and $(a^b)^c$ is obtained by multiplying together $c$ such products; writing all this out in terms of copies of $a$ means that $b\times c$ such copies have been multiplied together. Now you can see also why this is not equal to $a^{(b^c)}$ which is obtained by multiplying together $b^c$ copies of $a$.

Finally the fact that $(a^b)^c=a^{b\times c}$ explains why the convention is that exponentiation notation associates to the right: both $a^{(b^c)}$ and $(a^b)^c$ are useful expressions, but since the latter can be more easily written as $a^{bc}$, one might as well reserve $a^{b^c}$ to stand for the former, which has no such easy alternative. It might seem that $a^{b^c}$ is such an enormous product that is unlikely to be useful; it is however encountered surprisingly often in some contexts.


Note that $(2^2)^3=4^3=64$, whereas $2^{(2^3)}=2^8=256$ (and incidentally $2^{2\cdot 3}=2^6=64$).

  • $\begingroup$ +1 great, clean and nicely done! $\endgroup$ – doniyor Nov 18 '12 at 15:21
  • 2
    $\begingroup$ However I would need to subtract some points for $2^8=128$ if I found this on a test. $\endgroup$ – Marc van Leeuwen Nov 18 '12 at 15:35
  • $\begingroup$ @MarcvanLeeuwen Argh, numbers $\endgroup$ – Hagen von Eitzen Nov 18 '12 at 15:58

First question,

Why isn't $a^{b^c}$ equal to $a^{bc}$

assuming the remaining questions already answered. It is a matter of convention. Since $a^{(b^c)} \ne (a^b)^c$ in general, when we write $a^{b^c}$ we need convention to say which of these two we mean. Since $(a^b)^c = a^{bc}$, there is already a short way to write that, so we use $a^{b^c}$ for the other one.

  • $\begingroup$ Can we take a numeric example for showing him what is the story. Particularly, when the conventions looks a bit hard to understand? $\endgroup$ – mrs Nov 18 '12 at 14:34

First concept: Multiplication is NOT the same as exponentiation.

The question you are asking is, basically, why is $3^2 $ not the same as $3 (2)$ or $3 \times 2$?

Now, if we just take a look at an example problem involving your question:$$3^{2^{3}} = 3^{8} \ne {\left(3^{2}\right)^3}\ne 3^{2\times 3}\ne3^6$$ Notice that $3^{2^3}=3^8$ since $2^3 = 8$.

You are asking why multiplication is not the same as exponentiation, essentially. My answer is that multiplication is pertaining to repeated addition whereas exponentiation is pertaining to repeated multiplication.

If we look at fundamentals, $(a^b)^c$ is multiplying $a^b$ to itself $c$ times. Now if we multiply $a^b$ to itself an arbitrary number of times, then we are adding the power (here, $b$) to itself the same number of times. If you add something to itself some number of times, then you just multiply it by the number of times you are adding something to itself. Example: $$(2^3)^2 = 2^3 \times 2^3 = 2^{3 + 3} = 2^6 = 2^{3 \cdot 2}$$

And... $a^{b^c}$ means that you are multiplying $b$ to itself $c$ number of times.

For casual readers: skip to this part. In $(a^b)^c$, we multiply $b$ to $c$... and in $a^{b^c}$, we raise $b$ to the power $c$. As I have clearly repeated, exponentiation is not the same as addition is not the same as exponentiation. In other words, $bc \ne b^c \ne b + c$ always.

  • $\begingroup$ +1 here too: much cleaner example than I chose (now deleted)! Good for having pointed out explicitly that we are dealing with exponentiation of exponent, vs. multiplication in an exponent. $\endgroup$ – Namaste Nov 18 '12 at 14:47
  • $\begingroup$ @amWhy: Heh... more improvement now. $\endgroup$ – Parth Kohli Nov 18 '12 at 14:49
  • $\begingroup$ I think you mean, in your second to last sentence, that exponentiation is not the same as multiplication. (No worries, I understood what you were saying...)! $\endgroup$ – Namaste Nov 18 '12 at 14:55

Without parentheses: addition, subtraction, multiplication, and division should be evaluated from left to right. Like: $1 + 3 + 5 = 4 + 5 = 9$.

This rule does NOT apply for exponentiation, without parentheses, for exponentiation, we'll go from RIGHT to LEFT.

For example, to evaluate: $2^{2^3}$, we must evaluate $2^3 = 8$ first, so: $2^{2^3} = 2^8 = 256$.

So, without any parentheses, $a^{b^c}$ is the same as $a^{\left(b^c\right)}$, since we must go from RIGHT to LEFT.

For your second problem, why ${(a^b)}^c = a^{b.c}$.


If we take the sum of some number $a$ for $n$ times, we'll have multiplication, i.e $a \times n = \underbrace{a + a + a + ... + a}_{n \mbox { times}}$.

If we multiply some number $a$ for $n$ times, we'll have exponentiation, i.e $a ^ n = \underbrace{a \times a \times a \times ... \times a}_{n \mbox { times}}$.


  • $2 \times 3 = 2 + 2 + 2 = 6$
  • $2 ^ 3 = 2 \times 2 \times 2 = 4 \times 2 = 8$
  • $3 \times 4 = 3 + 3 + 3 + 3 = 12$
  • $3 ^ 4 = 3 \times 3 \times 3 \times 3 = 9 \times 3 \times 3 = 27 \times 3 = 81$


  • $a ^ m \times a ^ n = a^{m+n}$


$a ^ m \times a ^ n = \underbrace{\underbrace{a \times a \times a \times ... \times a}_{m \mbox { times}} \times \underbrace{a \times a \times a \times ... \times a}_{n \mbox { times}}}_{m + n \mbox { times}} = a^{m+n}$.
It's like 2 apples together with 3 apples becomes 2 + 3 = 5 apples. You have $n$ copies of $a$ together with another $m$ copies of $a$, you'll get $m + n$ copies of $a$.

  • $(a ^ m)^n = a^{m.n}$


$(a^m)^n$ basically means that you take the result of $a^m$, then raise the whole stuff to the power of $n$, or in other words, multiply $n$ copies of it together.

$(a^m)^n = \underbrace{a^m \times a^m \times ... \times a^m}_{n \mbox{ times}}$

Now, think of 5 groups of apples, such that that each group has exactly 2 apples. So there'll be a total of 2 x 5 = 10 apples. Each $a^m$ has $m$ copies of $a$, and there are $n$ copies of $a^m$, or in other words, there are $n$ groups, in which each group has $m$ copies of $a$. So there'll be a total of $m \times n$ copies of $a$. So:

$(a^m)^n = \underbrace{\underbrace{a \times a \times ... \times a}_{m \mbox{ times}} \times \underbrace{a \times a \times ... \times a}_{m \mbox{ times}} \times ... \times \underbrace{a \times a \times ... \times a}_{m \mbox{ times}}}_{n \mbox { times}} = a^{m.n}$.

  • $\begingroup$ good job by giving details! $\endgroup$ – doniyor Nov 18 '12 at 15:26

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